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Description: The Free High School Science Texts:
Textbooks for High School Students
Studying the Sciences
Mathematics
Grade 12
Copyright c 2007 “Free High School Science Texts” Permission is granted to c...

The Free High School Science Texts:
Textbooks for High School Students
Studying the Sciences
Mathematics
Grade 12
Copyright c 2007 “Free High School Science Texts” Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front- Cover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled “GNU Free Documentation License”
Webpage: http://www.fhsst.org/

FHSST Authors

The Free High School Science Texts: Textbooks for High School Students Studying the Sciences Mathematics Grades 10 - 12

Version 0 September 17, 2008

ii

iii Copyright 2007 “Free High School Science Texts” Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no FrontCover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled “GNU Free Documentation License”.

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Logarithms - Grade 12
In mathematics many ideas are related. We saw that addition and subtraction are related and that multiplication and division are related. Similarly, exponentials and logarithms are related. Logarithms, commonly referred to as logs, are the inverse of exponentials. The logarithm of a number x in the base a is deﬁned as the number n such that an = x. So, if an = x, then: loga (x) = n (35.1)

Extension: Inverse Function When we say “inverse function” we mean that the answer becomes the question and the question becomes the answer. For example, in the equation ab = x the “question” is “what is a raised to the power b.” The answer is “x.” The inverse function would be loga x = b or “by what power must we raise a to obtain x.” The answer is “b.” The mathematical symbol for logarithm is loga (x) and it is read “log to the base a of x”. For example, log10 (100) is “log to the base 10 of 100”.

The logarithm of a number is the index to which the base must be raised to give that number. From the ﬁrst example of the activity log2 (4) (read log to the base 2 of 4) means the power of 2 that will give 4. Therefore, log2 (4) = 2 (35.2) The index-form is then 22 = 4 and the logarithmic-form is log2 4 = 2. 445

Logarithms, like exponentials, also have a base and log2 (2) is not the same as log10 (2). We generally use the “common” base, 10, or the natural base, e. The number e is an irrational number between 2.71 and 2.72. It comes up surprisingly often in Mathematics, but for now suﬃce it to say that it is one of the two common bases.

Extension: Natural Logarithm The natural logarithm (symbol ln) is widely used in the sciences. The natural logarithm is to the base e which is approximately 2.71828183.... e is like π and is another example of an irrational number.

While the notation log10 (x) and loge (x) may be used, log10 (x) is often written log(x) in Science and loge (x) is normally written as ln(x) in both Science and Mathematics. So, if you see the log symbol without a base, it means log10 . It is often necessary or convenient to convert a log from one base to another. An engineer might need an approximate solution to a log in a base for which he does not have a table or calculator function, or it may be algebraically convenient to have two logs in the same base. Logarithms can be changed from one base to another, by using the change of base formula: loga x = logb x logb a (35.3)

where b is any base you ﬁnd convenient. Normally a and b are known, therefore logb a is normally a known, if irrational, number. For example, change log2 12 in base 10 is: log2 12 = log10 12 log10 2

When the base is 10, we do not need to state it. From the work done up to now, it is also useful to summarise the following facts: 1. log 1 = 0 2. log 10 = 1 3. log 100 = 2 4. log 1000 = 3

35.6

Logarithm Law 3: loga (x · y) = loga (x) + loga (y)

The derivation of this law is a bit trickier than the ﬁrst two. Firstly, we need to relate x and y to the base a. So, assume that x = am and y = an . Then from Equation 35.1, we have that: loga (x) and This means that we can write: loga (x · y) = = = = loga (am · an ) loga (am+n ) loga (y) = m = n (35.10) (35.11)

The derivation of this law is identical to the derivation of Logarithm Law 3 and is left as an exercise.
10 For example, show that log( 100 ) = log 10 − log 100. Start with calculating the left hand side:

In grade 10 you solved some exponential equations by trial and error, because you did not know the great power of logarithms yet. Now it is much easier to solve these equations by using logarithms. For example to solve x in 25x = 50 correct to two decimal places you simply apply the following reasoning. If the LHS = RHS then the logarithm of the LHS must be equal to the logarithm of the RHS. By applying Law 5, you will be able to use your calculator to solve for x.

In general, the exponential equation should be simpliﬁed as much as possible. Then the aim is to make the unknown quantity (i.e. x) the subject of the equation. For example, the equation 2(x+2) = 1 is solved by moving all terms with the unknown to one side of the equation and taking all constants to the other side of the equation 2x · 22 2
x

Worked Example 159: Exponential Equation Question: Solve for x in 7 · 5(3x+3) = 35 Answer Step 1 : Identify the base with x as an exponent There are two possible bases: 5 and 7. x is an exponent of 5. Step 2 : Eliminate the base with no x In order to eliminate 7, divide both sides of the equation by 7 to give: 5(3x+3) = 5 Step 3 : Take the logarithm of both sides log(5(3x+3) ) = log(5) Step 4 : Apply the log laws to make x the subject of the equation. 453

Logarithms are part of a number of formulae used in the Physical Sciences. There are formulae that deal with earthquakes, with sound, and pH-levels to mention a few. To work out time periods is growth or decay, logs are used to solve the particular equation.

Worked Example 160: Using the growth formula Question: A city grows 5% every 2 years. How long will it take for the city to triple its size? Answer Step 1 : Use the formula A = P (1 + i)n Assume P = x, then A = 3x. For this example n represents a period of 2 years, therefore the n is halved for this purpose. Step 2 : Substitute information given into formula 3 log 3 n n = = = = (1,05) 2 n × log 1.05 (usinglaw5) 2 2 log 3 ÷ log 1,05 45,034
n

Step 3 : Final answer So it will take approximately 45 years for the population to triple in size.

454

CHAPTER 35. LOGARITHMS - GRADE 12

35.12

35.11.1

Exercises

1. The population of a certain bacteria is expected to grow exponentially at a rate of 15 % every hour. If the initial population is 5 000, how long will it take for the population to reach 100 000 ? 2. Plus Bank is oﬀering a savings account with an interest rate if 10 % per annum compounded monthly. You can aﬀord to save R 300 per month. How long will it take you to save up R 20 000 ? (Answer to the nearest rand)

Worked Example 161: Logs in Compound Interest Question: I have R12 000 to invest. I need the money to grow to at least R30 000. If it is invested at a compound interest rate of 13% per annum, for how long (in full years) does my investment need to grow ? Answer Step 1 : The formula to use A = P (1 + i)n Step 2 : Substitute and solve for n 30 000 < 1,13n n n > > > 12 000(1 + 0,13)n 5 2 log 2,5 log 2,5 ÷ log 1,13 7,4972....

n log 1,13 >

Step 3 : Determine the ﬁnal answer In this case we round up, because 7 years will not yet deliver the required R 30 000. The investment need to stay in the bank for at least 8 years.

17. √ Solve the following equation for x without the use of a calculator and using the fact that 10 ≈ 3,16 : 6 2 log(x + 1) = −1 log(x + 1) 18. Solve the following equation for x: 66x = 66 (Give answer correct to 2 decimal places.)

456

Chapter 36

Sequences and Series - Grade 12
36.1 Introduction

In this chapter we extend the arithmetic and quadratic sequences studied in earlier grades, to geometric sequences. We also look at series, which is the summing of the terms in a sequence.

36.2

Arithmetic Sequences

The simplest type of numerical sequence is an arithmetic sequence.

Deﬁnition: Arithmetic Sequence An arithmetic (or linear ) sequence is a sequence of numbers in which each new term is calculated by adding a constant value to the previous term

For example, 1,2,3,4,5,6, . . . is an arithmetic sequence because you add 1 to the current term to get the next term: ﬁrst term: second term: third term: . . . nth term: 1 2=1+1 3=2+1 n = (n − 1) + 1

Given a1 and the common diﬀerence, d, the entire set of numbers belonging to an arithmetic sequence can be generated.

Deﬁnition: Arithmetic Sequence An arithmetic (or linear ) sequence is a sequence of numbers in which each new term is calculated by adding a constant value to the previous term: an = an−1 + d where • an represents the new term, the nth -term, that is calculated; • an−1 represents the previous term, the (n − 1)th -term; • d represents some constant. (36.2)

Important: Arithmetic Sequences

A simple test for an arithmetic sequence is to check that the diﬀerence between consecutive terms is constant: a2 − a1 = a3 − a2 = an − an−1 = d (36.3) This is quite an important equation, and is the deﬁnitive test for an arithmetic sequence. If this condition does not hold, the sequence is not an arithmetic sequence.

Extension: Plotting a graph of terms in an arithmetic sequence Plotting a graph of the terms of sequence sometimes helps in determining the type of sequence involved. For an arithmetic sequence, plotting an vs. n results in: 458

Deﬁnition: Geometric Sequences A geometric sequence is a sequence in which every number in the sequence is equal to the previous number in the sequence, multiplied by a constant number.

This means that the ratio between consecutive numbers in the geometric sequence is a constant. We will explain what we mean by ratio after looking at the following example.

36.3.1

Example - A Flu Epidemic

Extension: What is inﬂuenza? Inﬂuenza (commonly called “the ﬂu”) is caused by the inﬂuenza virus, which infects the respiratory tract (nose, throat, lungs). It can cause mild to severe illness that most of us get during winter time. The main way that the inﬂuenza virus is spread is from person to person in respiratory droplets of coughs and sneezes. (This is called “droplet spread”.) This can happen when droplets from a cough or sneeze of an infected person are propelled (generally, up to a metre) through the air and deposited on the mouth or nose of people nearby. It is good practise to cover your mouth when you cough or sneeze so as not to infect others around you when you have the ﬂu.

Assume that you have the ﬂu virus, and you forgot to cover your mouth when two friends came to visit while you were sick in bed. They leave, and the next day they also have the ﬂu. Let’s assume that they in turn spread the virus to two of their friends by the same droplet spread the following day. Assuming this pattern continues and each sick person infects 2 other friends, we can represent these events in the following manner: Again we can tabulate the events and formulate an equation for the general case: 459

The above table represents the number of newly-infected people after n days since you ﬁrst infected your 2 friends. You sneeze and the virus is carried over to 2 people who start the chain (a1 = 2). The next day, each one then infects 2 of their friends. Now 4 people are newly-infected. Each of them infects 2 people the third day, and 8 people are infected, and so on. These events can be written as a geometric sequence: 2; 4; 8; 16; 32; . . . Note the common factor (2) between the events. Recall from the linear arithmetic sequence how the common diﬀerence between terms were established. In the geometric sequence we can determine the common ratio, r, by a3 a2 = =r a1 a2 Or, more general, an =r an−1 (36.4)

From the above example we know a1 = 2 and r = 2, and we have seen from the table that the nth -term is given by an = 2 × 2n−1 . Thus, in general, an = a1 · rn−1 where a1 is the ﬁrst term and r is called the common ratio. So, if we want to know how many people are newly-infected after 10 days, we need to work out a10 : an a10 = = = = = a1 · rn−1 2 × 210−1 2 × 29 2 × 512 1024 (36.6)

That is, after 10 days, there are 1 024 newly-infected people. Or, how many days would pass before 16 384 people become newly infected with the ﬂu virus? an = a1 · rn−1

1. What is the important characteristic of an arithmetic sequence? 461

36.4

CHAPTER 36. SEQUENCES AND SERIES - GRADE 12

2. Write down how you would go about ﬁnding the formula for the nth term of an arithmetic sequence? 3. A single square is made from 4 matchsticks. Two squares in a row needs 7 matchsticks and 3 squares in a row needs 10 matchsticks. Determine: A the ﬁrst term B the common diﬀerence C the formula for the general term D how many matchsticks are in a row of 25 squares

4. 5; x; y is an arithmetic sequence and 81; x; y is a geometric sequence. All terms in the sequences are integers. Calculate the values of x and y.

36.4

Recursive Formulae for Sequences

When discussing arithmetic and quadratic sequences, we noticed that the diﬀerence between two consecutive terms in the sequence could be written in a general way. For an arithmetic sequence, where a new term is calculated by taking the previous term and adding a constant value, d: an = an−1 + d The above equation is an example of a recursive equation since we can calculate the nth -term only by considering the previous term in the sequence. Compare this with equation (36.1), an = a1 + d · (n − 1) (36.7)

where one can directly calculate the nth -term of an arithmetic sequence without knowing previous terms. For quadratic sequences, we noticed the diﬀerence between consecutive terms is given by (??): an − an−1 = D · (n − 2) + d Therefore, we re-write the equation as an = an−1 + D · (n − 2) + d Using (36.5), the recursive equation for a geometric sequence is: an = r · an−1 (36.9) (36.8)

which is then a recursive equation for a quadratic sequence with common second diﬀerence, D.

Recursive equations are extremely powerful: you can work out every term in the series just by knowing previous terms. As you can see from the examples above, working out an using the previous term an−1 can be a much simpler computation than working out an from scratch using a general formula. This means that using a recursive formula when using a computer to work out a sequence would mean the computer would ﬁnish its calculations signiﬁcantly quicker.

The above sequence is called the Fibonacci sequence. Each new term is calculated by adding the previous two terms. Hence, we can write down the recursive equation: an = an−1 + an−2 (36.11)

36.5

Series

In this section we simply work on the concept of adding up the numbers belonging to arithmetic and geometric sequences. We call the sum of any sequence of numbers a series.

36.5.1

Some Basics

If we add up the terms of a sequence, we obtain what is called a series. If we only sum a ﬁnite amount of terms, we get a ﬁnite series. We use the symbol Sn to mean the sum of the ﬁrst n terms of a sequence {a1 ; a2 ; a3 ; . . . ; an }: S n = a1 + a2 + a3 + . . . + an For example, if we have the following sequence of numbers 1; 4; 9; 25; 36; 49; . . . and we wish to ﬁnd the sum of the ﬁrst 4 terms, then we write S4 = 1 + 4 + 9 + 25 = 39 The above is an example of a ﬁnite series since we are only summing 4 terms. If we sum inﬁnitely many terms of a sequence, we get an inﬁnite series: S ∞ = a1 + a2 + a3 + . . . (36.13) (36.12)

In the case of an inﬁnite series, the number of terms is unknown and simply increases to ∞.

36.5.2

Sigma Notation

In this section we introduce a notation that will make our lives a little easier. 463

36.5

CHAPTER 36. SEQUENCES AND SERIES - GRADE 12

A sum may be written out using the summation symbol . This symbol is sigma, which is the capital letter “S” in the Greek alphabet. It indicates that you must sum the expression to the right of it:
n

ai = am + am+1 + . . . + an−1 + an
i=m

(36.14)

where • i is the index of the sum; • m is the lower bound (or start index), shown below the summation symbol; • n is the upper bound (or end index), shown above the summation symbol; • ai are the terms of a sequence. The index i is increased from m to n in steps of 1. If we are summing from n = 1 (which implies summing from the ﬁrst term in a sequence), then we can use either Sn - or -notation since they mean the same thing:
n

Sn =
i=1

ai = a1 + a2 + . . . + an

(36.15)

For example, in the following sum,
5

i
i=1

we have to add together all the terms in the sequence ai = i from i = 1 up until i = 5:
5

Remember that an arithmetic sequence is a set of numbers, such that the diﬀerence between any term and the previous term is a constant number, d, called the constant diﬀerence: an = a1 + d (n − 1) where • n is the index of the sequence; • an is the nth -term of the sequence; • a1 is the ﬁrst term; • d is the common diﬀerence. When we sum a ﬁnite number of terms in an arithmetic sequence, we get a ﬁnite arithmetic series. The simplest arithmetic sequence is when a1 = 1 and d = 0 in the general form (36.18); in other words all the terms in the sequence are 1: ai = = a1 + d (i − 1) 1 + 0 · (i − 1) (36.18)

= {ai } =
n n

1 {1; 1; 1; 1; 1; . . .}

If we wish to sum this sequence from i = 1 to any positive integer n, we would write ai =
i=1 i=1

1 = 1 + 1 + 1 + ...+ 1 465

(n times)

36.6

CHAPTER 36. SEQUENCES AND SERIES - GRADE 12

Since all the terms are equal to 1, it means that if we sum to n we will be adding n-number of 1’s together, which is simply equal to n:
n

If we wish to sum this sequence from i = 1 to any positive integer n, we would write i = 1 + 2 + 3 + ...+ n
i=1

(36.20)

This is an equation with a very important solution as it gives the answer to the sum of positive integers.

teresting Mathematician, Karl Friedrich Gauss, discovered this proof when he was only Interesting Fact Fact 8 years old. His teacher had decided to give his class a problem which would distract them for the entire day by asking them to add all the numbers from 1 to 100. Young Karl realised how to do this almost instantaneously and shocked the teacher with the correct answer, 5050.

or, more sensibly, we could use equation (36.25) noting that a1 = 3, d = 7 and n = 20 so that
20

S20

= = =

i=1 20 2 [2

[3 + 7 (i − 1)] · 3 + 7 (20 − 1)]

1390

In this example, it is clear that using equation (36.25) is beneﬁcial.

36.6.2

Exercises
n (7n + 15). 2

1. The sum to n terms of an arithmetic series is Sn =

A How many terms of the series must be added to give a sum of 425? B Determine the 6th term of the series. 2. The sum of an arithmetic series is 100 times its ﬁrst term, while the last term is 9 times the ﬁrst term. Calculate the number of terms in the series if the ﬁrst term is not equal to zero. 467

36.7

CHAPTER 36. SEQUENCES AND SERIES - GRADE 12

3. The common diﬀerence of an arithmetic series is 3. Calculate the values of n for which the nth term of the series is 93, and the sum of the ﬁrst n terms is 975. 4. The sum of n terms of an arithmetic series is 5n2 − 11n for all values of n. Determine the common diﬀerence. 5. The sum of an arithmetic series is 100 times the value of its ﬁrst term, while the last term is 9 times the ﬁrst term. Calculate the number of terms in the series if the ﬁrst term is not equal to zero. 6. The third term of an arithmetic sequence is -7 and the 7t h term is 9. Determine the sum of the ﬁrst 51 terms of the sequence. 7. Calculate the sum of the arithmetic series 4 + 7 + 10 + · · · + 901. 8. The common diﬀerence of an arithmetic series is 3. Calculate the values of n for which the nth term of the series is 93 and the sum of the ﬁrst n terms is 975.

36.7

Finite Squared Series

When we sum a ﬁnite number of terms in a quadratic sequence, we get a ﬁnite quadratic series. The general form of a quadratic series is quite complicated, so we will only look at the simple case when D = 2 and d = (a2 − a1 ) = 3 in the general form (???). This is the sequence of squares of the integers: ai {ai } = = = i2 {12 ; 22 ; 32 ; 42 ; 52 ; 62 ; . . .} {1; 4; 9; 16; 25; 36; . . .}
n

If we wish to sum this sequence and create a series, then we write Sn =
i=1

i 2 = 1 + 4 + 9 + . . . + n2

which can be written, in general, as
n

Sn =
i=1

i2 =

n (2n + 1)(n + 1) 6

(36.26)

The proof for equation (36.26) can be found under the Advanced block that follows:

Extension: Derivation of the Finite Squared Series We will now prove the formula for the ﬁnite squared series:
n

When we sum a known number of terms in a geometric sequence, we get a ﬁnite geometric series. We know from (??) that we can write out each term of a geometric sequence in the general form: an = a1 · rn−1 (36.27) where • n is the index of the sequence; • an is the nth -term of the sequence; • a1 is the ﬁrst term; • r is the common ratio (the ratio of any term to the previous term). By simply adding together the ﬁrst n terms, we are actually writing out the series Sn = a1 + a1 r + a1 r2 + . . . + a1 rn−2 + a1 rn−1 We may multiply the above equation by r on both sides, giving us rSn = a1 r + a1 r2 + a1 r3 + . . . + a1 rn−1 + a1 rn 469 (36.29) (36.28)

36.8

CHAPTER 36. SEQUENCES AND SERIES - GRADE 12

You may notice that all the terms on the right side of (36.28) and (36.29) are the same, except the ﬁrst and last terms. If we subtract (36.28) from (36.29), we are left with just rSn − Sn = a1 rn − a1 Sn (r − 1) = a1 (rn − 1) Dividing by (r − 1) on both sides, we arrive at the general form of a geometric series:
n

Sn =
i=1

a1 · ri−1 =

a1 (rn − 1) r−1

(36.30)

36.8.1

Exercises
a + ar + ar2 + ... + arn−1 = a (1 − rn ) (1 − r)
3 2

1. Prove that

2. Find the sum of the ﬁrst 11 terms of the geometric series 6 + 3 + 3. Show that the sum of the ﬁrst n terms of the geometric series
1 54 + 18 + 6 + ... + 5 ( 3 )n−1

+

3 4

+ ...

is given by 81 − 34−n . 4. The eighth term of a geometric sequence is 640. The third term is 20. Find the sum of the ﬁrst 7 terms.
n

5. Solve for n:
t=1

3 1 8 ( 2 )t = 15 4 .

6. The ratio between the sum of the ﬁrst three terms of a geometric series and the sum of the 4th -, 5th − and 6th -terms of the same series is 8 : 27. Determine the common ratio and the ﬁrst 2 terms if the third term is 8. 7. Given the geometric series: 2 · (5)5 + 2 · (5)4 + 2 · (5)3 + . . . A Show that the series converges B Calculate the sum to inﬁnity of the series C Calculate the sum of the ﬁrst 8 terms of the series, correct to two decimal places. D Determine
∞ n=9

2 · 56−n

correct to two decimal places using previously calculated results. 8. Given the geometric sequence 1; −3; 9; . . . determine: A The 8th term of the sequence B The sum of the ﬁrst 8 terms of the sequence. 9. Determine:
4 n=1

3 · 2n−1

470

CHAPTER 36. SEQUENCES AND SERIES - GRADE 12

36.9

36.9

Inﬁnite Series

Thus far we have been working only with ﬁnite sums, meaning that whenever we determined the sum of a series, we only considered the sum of the ﬁrst n terms. In this section, we consider what happens when we add inﬁnitely many terms together. You might think that this is a silly question - surely the answer will be ∞ when one sums inﬁnitely many numbers, no matter how small they are? The surprising answer is that in some cases one will reach ∞ (like when you try to add all the positive integers together), but in some cases one will get a ﬁnite answer. If you don’t believe this, try doing the following sum, a geometric series, on your calculator or computer: 1 1 1 1 1 2 + 4 + 8 + 16 + 32 + . . . You might think that if you keep adding more and more terms you will eventually get larger and larger numbers, but in fact you won’t even get past 1 - try it and see for yourself! We denote the sum of an inﬁnite number of terms of a sequence by S∞ =
∞ i=1

ai

When we sum the terms of a series, and the answer we get after each summation gets closer and closer to some number, we say that the series converges. If a series does not converge, then we say that it diverges.

36.9.1

Inﬁnite Geometric Series

There is a simple test for knowing instantly which geometric series converges and which diverges. When r, the common ratio, is strictly between -1 and 1, i.e. −1 < r < 1, the inﬁnite series will converge, otherwise it will diverge. There is also a formula for working out what the series converges to. Let’s start oﬀ with formula (36.30) for the ﬁnite geometric series:
n

calculate the smallest value of n for which the sum of the ﬁrst n terms is greater than 80.99. 7. Determine the value of
∞ k=1 1 12( 5 )k−1 .

8. A new soccer competition requires each of 8 teams to play every other team once. A Calculate the total number of matches to be played in the competition. 472

CHAPTER 36. SEQUENCES AND SERIES - GRADE 12

36.10

B If each of n teams played each other once, determine a formula for the total number of matches in terms of n. 9. The midpoints of the sides of square with length equal to 4 units are joined to form a new square. The process is repeated indeﬁnitely. Calculate the sum of the areas of all the squares so formed. 10. Thembi worked part-time to buy a Mathematics book which cost R29,50. On 1 February she saved R1,60, and saves everyday 30 cents more than she saved the previous day. (So, on the second day, she saved R1,90, and so on.) After how many days did she have enough money to buy the book? 11. Consider the geometric series: 5 + 21 + 11 + . . . 2 4 A If A is the sum to inﬁnity and B is the sum of the ﬁrst n terms, write down the value of: i. A ii. B in terms of n. B For which values of n is (A − B) <
1 24 ?

12. A certain plant reaches a height of 118 mm after one year under ideal conditions in a greenhouse. During the next year, the height increases by 12 mm. In each successive year, 5 the height increases by 8 of the previous year’s growth. Show that the plant will never reach a height of more than 150 mm.
n

13. Calculate the value of n if
a=1

(20 − 4a) = −20.

14. Michael saved R400 during the ﬁrst month of his working life. In each subsequent month, he saved 10% more than what he had saved in the previous month. A How much did he save in the 7th working month? B How much did he save all together in his ﬁrst 12 working months? C In which month of his working life did he save more than R1,500 for the ﬁrst time? 15. A man was injured in an accident at work. He receives a disability grant of R4,800 in the ﬁrst year. This grant increases with a ﬁxed amount each year. A What is the annual increase if, over 20 years, he would have received a total of R143,500? B His initial annual expenditure is R2,600 and increases at a rate of R400 per year. After how many years does his expenses exceed his income? 16. The Cape Town High School wants to build a school hall and is busy with fundraising. Mr. Manuel, an ex-learner of the school and a successful politician, oﬀers to donate money to the school. Having enjoyed mathematics at school, he decides to donate an amount of money on the following basis. He sets a mathematical quiz with 20 questions. For the correct answer to the ﬁrst question (any learner may answer), the school will receive 1 cent, for a correct answer to the second question, the school will receive 2 cents, and so on. The donations 1, 2, 4, ... form a geometric sequence. Calculate (Give your answer to the nearest Rand) A The amount of money that the school will receive for the correct answer to the 20th question. B The total amount of money that the school will receive if all 20 questions are answered correctly. 17. The ﬁrst term of a geometric sequence is 9, and the ratio of the sum of the ﬁrst eight terms to the sum of the ﬁrst four terms is 97 : 81. Find the ﬁrst three terms of the sequence, if it is given that all the terms are positive. 18. (k − 4); (k + 1); m; 5k is a set of numbers, the ﬁrst three of which form an arithmetic sequence, and the last three a geometric sequence. Find k and m if both are positive. 473

C Determine the 10th term of this sequence correct to one decimal place. 20. The second and fourth terms of a convergent geometric series are 36 and 16, respectively. Find the sum to inﬁnity of this series, if all its terms are positive.
5

A If the pattern continues, ﬁnd the number of letters in the column containing M’s. B If the total number of letters in the pattern is 361, which letter will the last column consist of. 31. The following question was asked in a test: Find the value of 22005 + 22005 . Here are some of the students’ answers: A Megansaid the answer is 42005 . B Stefan wrote down 24010 . C Nina thinks it is 22006 . D Annatte gave the answer 22005×2005 . Who is correct? (“None of them” is also a possibility.) 32. Find the pattern and hence calculate: 1 − 2 + 3 − 4 + 5 − 6 . . . + 677 − 678 + . . . − 1000
∞

33. Determine A x=−

(x + 2)p , if it exists, when

p=1

5 2 B x = −5
∞ i=1

34. Calculate:

5 · 4−i 475

36.10

CHAPTER 36. SEQUENCES AND SERIES - GRADE 12

35. The sum of the ﬁrst p terms of a sequence is p (p + 1). Find the 10th term. 36. The powers of 2 are removed from the set of positive integers 1; 2; 3; 4; 5; 6; . . . ; 1998; 1999; 2000 Find the sum of remaining integers. 37. A shrub of height 110 cm is planted. At the end of the ﬁrst year, the shrub is 120 cm tall. Thereafter, the growth of the shrub each year is half of its growth in the previous year. Show that the height of the shrub will never exceed 130 cm.

476

Chapter 37

Finance - Grade 12
37.1 Introduction

In earlier grades simple interest and compound interest were studied, together with the concept of depreciation. Nominal and eﬀective interest rates were also described. Since this chapter expands on earlier work, it would be best if you revised the work done in Chapters 8 and 21. If you master the techniques in this chapter, when you start working and earning you will be able to apply the techniques in this chapter to critically assess how to invest your money. And when you are looking at applying for a bond from a bank to buy a home, you will conﬁdently be able to get out the calculator and work out with amazement how much you could actually save by making additional repayments. Indeed, this chapter will provide you with the fundamental concepts you will need to conﬁdently manage your ﬁnances and with some successful investing, sit back on your yacht and enjoy the millionaire lifestyle.

37.2

Finding the Length of the Investment or Loan

In Grade 11, we used the formula A = P (1 + i)n to determine the term of the investment or loan, by trial and error. In other words, if we know what the starting sum of money is and what it grows to, and if we know what interest rate applies - then we can work out how long the money needs to be invested for all those other numbers to tie up. Now, that you have learnt about logarithms, you are ready to work out the proper algebraic solution. If you need to remind yourself how logarithms work, go to Chapter 35 (on page 445). The basic ﬁnance equation is: A = P · (1 + i)n If you don’t know what A, P , i and n represent, then you should deﬁnitely revise the work from Chapters 8 and 21. Solving for n: A = (1 + i)n = log((1 + i)n ) = n log(1 + i) = n = P (1 + i)n (A/P ) log(A/P ) log(A/P ) log(A/P )/ log(1 + i)

Remember, you do not have to memorise this formula. It is very easy to derive any time you need it. It is simply a matter of writing down what you have, deciding what you need, and solving for that variable.

By this stage, you know how to do calculations such as ”If I want R1 000 in 3 years’ time, how much do I need to invest now at 10% ?” But what if we extend this as follows: If I want R1 000 next year and R1 000 the year after that and R1 000 after three years ... how much do I need to put into a bank account earning 10% p.a. right now to be able to aﬀord that?” The obvious way of working that out is to work out how much you need now to aﬀord the payments individually and sum them. We’ll work out how much is needed now to aﬀord the payment of R1 000 in a year (= R1 000 × (1,10)−1 = R909,0909), the amount needed now for the following year’s R1 000 (= R1 000 × (1,10)−2 = R826,4463) and the amount needed now for the R1 000 after 3 years (= R1 000 × (1,10)−3 = R751,3148). Add these together gives you the amount needed to aﬀord all three payments and you get R2486,85. So, if you put R2486,85 into a 10% bank account now, you will be able to draw out R1 000 in a year, R1 000 a year after that, and R1 000 a year after that - and your bank account will come down to R0. You would have had exactly the right amount of money to do that (obviously!). You can check this as follows: 478

CHAPTER 37. FINANCE - GRADE 12 Amount Amount Amount Amount Amount Amount Amount at Time 0 (i.e. Now) at Time 1 (i.e. a year later) after the R1 000 at Time 2 (i.e. a year later) after the R1 000 at Time 3 (i.e. a year later) after the R1 000 = = = = = = = R2486,85 R2735,54 R1735,54 R1909,09 R909,09 R1 000 R0

Perfect! Of course, for only three years, that was not too bad. But what if I asked you how much you needed to put into a bank account now, to be able to aﬀord R100 a month for the next 15 years. If you used the above approach you would still get the right answer, but it would take you weeks! There is - I’m sure you guessed - an easier way! This section will focus on describing how to work with: • annuities - a ﬁxed sum payable each year or each month either to provide a pre-determined sum at the end of a number of years or months (referred to as a future value annuity) or a ﬁxed amount paid each year or each month to repay (amortise) a loan (referred to as a present value annuity). • bond repayments - a ﬁxed sum payable at regular intervals to pay oﬀ a loan. This is an example of a present value annuity. • sinking funds - an accounting term for cash set aside for a particular purpose and invested so that the correct amount of money will be available when it is needed. This is an example of a future value annuity

37.3.1

Sequences and Series

Before we progress, you need to go back and read Chapter 36 (from page 457) to revise sequences and series. In summary, if you have a series of n terms in total which looks like this: a + ar + ar2 + ... + arn−1 = a[1 + r + r2 + ...rn−1 ] this can be simpliﬁed as: a(rn − 1) r−1 a(1 − rn ) 1−r useful when r > 1 useful when 0 ≤ r < 1

37.3.2

Present Values of a series of Payments

So having reviewed the mathematics of Sequences and Series, you might be wondering how this is meant to have any practical purpose! Given that we are in the ﬁnance section, you would be right to guess that there must be some ﬁnancial use to all this Here is an example which happens in many people’s lives - so you know you are learning something practical Let us say you would like to buy a property for R300 000, so you go to the bank to apply for a mortgage bond. The bank wants it to be repaid by annually payments for the next 20 years, starting at end of this year. They will charge you 15% per annum. At the end of the 20 years the bank would have received back the total amount you borrowed together with all the interest they have earned from lending you the money. You would obviously want to work out what the annual repayment is going to be! Let X be the annual repayment, i is the interest rate, and M is the amount of the mortgage bond you will be taking out. Time lines are particularly useful tools for visualizing the series of payments for calculations, and we can represent these payments on a time line as: 479

37.3

CHAPTER 37. FINANCE - GRADE 12

X 0 1

X 2

X 18

X 19

X 20

Cash Flows Time

Figure 37.1: Time Line for an annuity (in arrears) of X for n periods.

The present value of all the payments (which includes interest) must equate to the (present) value of the mortgage loan amount. Mathematically, you can write this as: M = X(1 + i)−1 + X(1 + i)−2 + X(1 + i)−3 + ... + X(1 + i)−20 The painful way of solving this problem would be to do the calculation for each of the terms above - which is 20 diﬀerent calculations. Not only would you probably get bored along the way, but you are also likely to make a mistake. Naturally, there is a simpler way of doing this! You can rewrite the above equation as follows:

M = X(v 1 + v 2 + v 3 + ... + v 20 ) where v = (1 + i)−1 = 1/(1 + i) Of course, you do not have to use the method of substitution to solve this. We just ﬁnd this a useful method because you can get rid of the negative exponents - which can be quite confusing! As an exercise - to show you are a real ﬁnancial whizz - try to solve this without substitution. It is actually quite easy. Now, the item in square brackets is the sum of a geometric sequence, as discussion in section 36. This can be re-written as follows, using what we know from Chapter 36 of this text book: v 1 + v 2 + v 3 + ... + v n = v(1 + v + v 2 + ... + v n−1 ) 1 − vn ) = v( 1−v 1 − vn = 1/v − 1 1 − (1 + i)−n = i

Note that we took out a common factor of v before using the formula for the geometric sequence. So we can write: M = X[ This can be re-written: X= (1 − (1 + i)−n ) ] i M
−n [ (1−(1+i) ) ] i

So, this formula is useful if you know the amount of the mortgage bond you need and want to work out the repayment, or if you know how big a repayment you can aﬀord and want to see what property you can buy. For example, if I want to buy a house for R300 000 over 20 years, and the bank is going to 480

= R47 928,44 This means, each year for the next 20 years, I need to pay the bank R47 928,44 per year before I have paid oﬀ the mortgage bond. On the other hand, if I know I will only have R30 000 a year to repay my bond, then how big a house can I buy? That is easy ....

So, for R30 000 a year for 20 years, I can aﬀord to buy a house of R187 800 (rounded to the nearest hundred). The bad news is that R187 800 does not come close to the R300 000 you wanted to buy! The good news is that you do not have to memorise this formula. In fact , when you answer questions like this in an exam, you will be expected to start from the beginning - writing out the opening equation in full, showing that it is the sum of a geometric sequence, deriving the answer, and then coming up with the correct numerical answer.

Worked Example 163: Monthly mortgage repayments Question: Sam is looking to buy his ﬁrst ﬂat, and has R15 000 in cash savings which he will use as a deposit. He has viewed a ﬂat which is on the market for R250 000, and he would like to work out how much the monthly repayments would be. He will be taking out a 30 year mortgage with monthly repayments. The annual interest rate is 11%. Answer Step 1 : Determine what is given and what is needed The following is given: • Deposit amount = R15 000 • Price of ﬂat = R250 000 • interest rate, i = 11% We are required to ﬁnd the monthly repayment for a 30-year mortgage. Step 2 : Determine how to approach the problem We know that: M X = (1−(1+i)−n ) ] [ i . In order to use this equation, we need to calculate M , the amount of the mortgage bond, which is the purchase price of property less the deposit which Sam pays upfront. M = = R250 000 − R15 000 R235 000 481

37.3

CHAPTER 37. FINANCE - GRADE 12 Now because we are considering monthly repayments, but we have been given an annual interest rate, we need to convert this to a monthly interest rate, i12. (If you are not clear on this, go back and revise section 21.8.) (1 + i12)12
12

=

(1 + i) 1,11 0,873459%

(1 + i12) = i12 =

We know that the mortgage bond is for 30 years, which equates to 360 months. Step 3 : Solve the problem Now it is easy, we can just plug the numbers in the formula, but do not forget that you can always deduce the formula from ﬁrst principles as well! M
−n [ (1−(1+i) ) ] i

X

= = =

R235 000
−360 ) ] [ (1−(1.00876459) 0,008734594

R2 146,39

Step 4 : Write the ﬁnal answer That means that to buy a house for R300 000, after Sam pays a R15 000 deposit, he will make repayments to the bank each month for the next 30 years equal to R2 146,39.

Worked Example 164: Monthly mortgage repayments Question: You are considering purchasing a ﬂat for R200 000 and the bank’s mortgage rate is currently 9% per annum payable monthly. You have savings of R10 000 which you intend to use for a deposit. How much would your monthly mortgage payment be if you were considering a mortgage over 20 years. Answer Step 1 : Determine what is given and what is required The following is given: • Deposit amount = R10 000 • Price of ﬂat = R200 000 • interest rate, i = 9% We are required to ﬁnd the monthly repayment for a 20-year mortgage. Step 2 : Determine how to approach the problem We are consider monthly mortgage repayments, so it makes sense to use months as our time period. The interest rate was quoted as 9% per annum payable monthly, which means that the monthly eﬀective rate = 9%/12 = 0,75% per month. Once we have converted 20 years into 240 months, we are ready to do the calculations! First we need to calculate M , the amount of the mortgage bond, which is the purchase price of property less the deposit which Sam pays up-front. M = R200 000 − R10 000 = R190 000 482

But it is clearly much easier to use our formula that work out 240 factors and add them all up! Step 3 : Solve the problem 1 − (1 + 0,75%)−240 = 0,75% X × 111,14495 = X =

X×

R190 000 R190 000 R1 709,48

Step 4 : Write the ﬁnal answer So to repay a R190 000 mortgage over 20 years, at 9% interest payable monthly, will cost you R1 709,48 per month for 240 months.

Show me the money Now that you’ve done the calculations for the worked example and know what the monthly repayments are, you can work out some surprising ﬁgures. For example, R1 709,48 per month for 240 month makes for a total of R410 275,20 (=R1 709,48 × 240). That is more than double the amount that you borrowed! This seems like a lot. However, now that you’ve studied the eﬀects of time (and interest) on money, you should know that this amount is somewhat meaningless. The value of money is dependant on its timing. Nonetheless, you might not be particularly happy to sit back for 20 years making your R1 709,48 mortgage payment every month knowing that half the money you are paying are going toward interest. But there is a way to avoid those heavy interest charges. It can be done for less than R300 extra every month... So our payment is now R2 000. The interest rate is still 9% per annum payable monthly (0,75% per month), and our principal amount borrowed is R190 000. Making this higher repayment amount every month, how long will it take to pay oﬀ the mortgage? The present value of the stream of payments must be equal to R190 000 (the present value of the borrowed amount). So we need to solve for n in: R2 000 × [1 − (1 + 0,75%)−n ]/0,75% = 1 − (1 + 0,75%) log(1 + 0,75%)−n
−n

R190 000 (R190 000/2 000) × 0,75% log[(1 − (R190 000/R2 000) × 0,75%]

= =

−n × log(1 + 0,75%) = −n × 0,007472 = n = =

log[(1 − (R190 000/R2 000) × 0,75%] −1,2465 166,8 months 13,9 years

So the mortgage will be completely repaid in less than 14 years, and you would have made a total payment of 166,8× R2 000 = R333 600. Can you see what is happened? Making regular payments of R2 000 instead of the required R1,709,48, you will have saved R76 675,20 (= R410 275,20 - R333 600) in interest, and yet you have only paid an additional amount of R290,52 for 166,8 months, or R48 458,74. You surely 483

37.3

CHAPTER 37. FINANCE - GRADE 12

know by now that the diﬀerence between the additional R48 458,74 that you have paid and the R76 675,20 interest that you have saved is attributable to, yes, you have got it, compound interest!

37.3.3

Future Value of a series of Payments

In the same way that when we have a single payment, we can calculate a present value or a future value - we can also do that when we have a series of payments. In the above section, we had a few payments, and we wanted to know what they are worth now - so we calculated present values. But the other possible situation is that we want to look at the future value of a series of payments. Maybe you want to save up for a car, which will cost R45 000 - and you would like to buy it in 2 years time. You have a savings account which pays interest of 12% per annum. You need to work out how much to put into your bank account now, and then again each month for 2 years, until you are ready to buy the car. Can you see the diﬀerence between this example and the ones at the start of the chapter where we were only making a single payment into the bank account - whereas now we are making a series of payments into the same account? This is a sinking fund. So, using our usual notation, let us write out the answer. Make sure you agree how we come up with this. Because we are making monthly payments, everything needs to be in months. So let A be the closing balance you need to buy a car, P is how much you need to pay into the bank account each month, and i12 is the monthly interest rate. (Careful - because 12% is the annual interest rate, so we will need to work out later what the month interest rate is!) A = P (1 + i12)24 + P (1 + i12)23 + ... + P (1 + i12)1 Here are some important points to remember when deriving this formula: 1. We are calculating future values, so in this example we use (1 + i12)n and not (1 + i12)−n . Check back to the start of the chapter is this is not obvious to you by now. 2. If you draw a timeline you will see that the time between the ﬁrst payment and when you buy the car is 24 months, which is why we use 24 in the ﬁrst exponent. 3. Again, looking at the timeline, you can see that the 24th payment is being made one month before you buy the car - which is why the last exponent is a 1. 4. Always check that you have got the right number of payments in the equation. Check right now that you agree that there are 24 terms in the formula above. So, now that we have the right starting point, let us simplify this equation: A = = P [(1 + i12)24 + (1 + i12)23 + . . . + (1 + i12)1 ] P [X 24 + X 23 + . . . + X 1 ] using X = (1 + i12)

Note that this time X has a positive exponent not a negative exponent, because we are doing future values. This is not a rule you have to memorise - you can see from the equation what the obvious choice of X should be. Let us reorder the terms: A = P [X 1 + X 2 + . . . + X 24 ] = P · X[1 + X + X 2 + . . . + X 2 3] This is just another sum of a geometric sequence, which as you know can be simpliﬁed as: A = = P · X[X n − 1]/((1 + i12) − 1) P · X[X n − 1]/i12 484

CHAPTER 37. FINANCE - GRADE 12

37.4

So if we want to use our numbers, we know that A = R45 000, n=24 (because we are looking at monthly payments, so there are 24 months involved) and i = 12% per annum. BUT (and it is a big but) we need a monthly interest rate. Do not forget that the trick is to keep the time periods and the interest rates in the same units - so if we have monthly payments, make sure you use a monthly interest rate! Using the formula from Section 21.8, we know that (1 + i) = (1 + i12)12 . So we can show that i12 = 0,0094888 = 0,94888%. Therefore, 45 000 = P = P (1,0094888)[(1,0094888)24 − 1]/0,0094888 1662,67

This means you need to invest R1 662,67 each month into that bank account to be able to pay for your car in 2 years time. There is another way of looking at this too - in terms of present values. We know that we need an amount of R45 000 in 24 months time, and at a monthly interest rate of 0,94888%, the present value of this amount is R35 873,72449. Now the question is what monthly amount at 0,94888% interest over 24 month has a present value of R35 873,72449? We have seen this before - it is just like the mortgage questions! So let us go ahead and see if we get to the same answer

P

= R35 873,72449[(1 − (1,0094888)−24)/0,0094888] = R1 662,67

= M/[(1 − (1 + i)−n )/i]

37.3.4

Exercises - Present and Future Values

1. You have taken out a mortgage bond for R875 000 to buy a ﬂat. The bond is for 30 years and the interest rate is 12% per annum payable monthly. A What is the monthly repayment on the bond? B How much interest will be paid in total over the 30 years? 2. How much money must be invested now to obtain regular annuity payments of R 5 500 per month for ﬁve years ? The money is invested at 11,1% p.a., compounded monthly. (Answer to the nearest hundred rand)

37.4

Investments and Loans

By now, you should be well equipped to perform calculations with compound interest. This section aims to allow you to use these valuable skills to critically analyse investment and load options that you will come across in your later life. This way, you will be able to make informed decisions on options presented to you. At this stage, you should understand the mathematical theory behind compound interest. However, the numerical implications of compound interest is often subtle and far from obvious. Recall the example in section ??FIXTHIS. For an extra payment of R290,52 a month, we could have paid oﬀ our loan in less than 14 years instead of 20 years. This provides a good illustration of the long term eﬀect of compound interest that is often surprising. In the following section, we’ll aim to explain the reason for drastic deduction in times it takes to repay the loan.

37.4.1

Loan Schedules

So far, we have been working out loan repayment amounts by taking all the payments and discounting them back to the present time. We are not considering the repayments individually. 485

37.4

CHAPTER 37. FINANCE - GRADE 12

Think about the time you make a repayment to the bank. There are numerous questions that could be raised: how much do you still owe them? Since you are paying oﬀ the loan, surely you must owe them less money, but how much less? We know that we’ll be paying interest on the money we still owe the bank. When exactly do we pay interest? How much interest are we paying? The answer to these questions lie in something called the load schedule. We will continue to use the example from section ??FIXTHIS. There is a loan amount of R190 000. We are paying it oﬀ over 20 years at an interest of 9% per annum payable monthly. We worked out that the repayments should be R1 709,48. Consider the ﬁrst payment of R1 709,48 one month into the loan. First, we can work out how much interest we owe the bank at this moment. We borrowed R190 000 a month ago, so we should owe:

I

= = =

M × i12 R190 000 × 0,75% R1 425

We are paying them R1 425 in interest. We calls this the interest component of the repayment. We are only paying oﬀ R1 709,48 - R1 425 = R284.48 of what we owe! This is called the capital component. That means we still owe R190 000 - R284,48 = R189 715,52. This is called the capital outstanding. Let’s see what happens at end of the second month. The amount of interest we need to pay is the interest on the capital outstanding.

I

= = =

M × i12

R189 715,52 × 0,75% R1 422,87

Since we don’t owe the bank as much as we did last time, we also owe a little less interest. The capital component of the repayment is now R1 709,48 - R1 422,87 = R286,61. The capital outstanding will be R189 715,52 - R286,61 = R189 428,91. This way, we can break each of our repayments down into an interest part and the part that goes towards paying oﬀ the loan. This is a simple and repetitive process. Table 37.1 is a table showing the breakdown of the ﬁrst 12 payments. This is called a loan schedule. Now, let’s see the same thing again, but with R2 000 being repaid each year. We expect the numbers to change. However, how much will they change by? As before, we owe R1 425 in interest in interest. After one month. However, we are paying R2 000 this time. That leaves R575 that goes towards paying oﬀ the capital outstanding, reducing it to R189 425. By the end of the second month, the interest owed is R1 420,69 (That’s R189 425×i12). Our R2 000 pays for that interest, and reduces the capital amount owed by R2 000 - R1 420,69 = R579,31. This reduces the amount outstanding to R188 845,69. Doing the same calculations as before yields a new loan schedule shown in Table 37.2. The important numbers to notice is the “Capital Component” column. Note that when we are paying oﬀ R2 000 a month as compared to R1 709,48 a month, this column more than doubles? In the beginning of paying oﬀ a loan, very little of our money is used to pay oﬀ the captital outstanding. Therefore, even a small incread in repayment amounts can signiﬁcantly increase the speed at which we are paying oﬀ the capital. Whatsmore, look at the amount we are still owing after one year (i.e. at time 12). When we were paying R1 709,48 a month, we still owe R186 441,84. However, if we increase the repayments to R2 000 a month, the amount outstanding decreases by over R3 000 to R182 808,14. This means we would have paid oﬀ over R7 000 in our ﬁrst year instead of less than R4 000. This 486

increased speed at which we are paying oﬀ the capital portion of the loan is what allows us to pay oﬀ the whole load in around 14 years instead of the original 20. Note however, the eﬀect of paying R2 000 instead of R1 709,48 is more signiﬁcant in be beginning of the loan than near the end of the loan. It is noted that in this instance, by paying slightly more than what the bank would ask you to pay, you can pay oﬀ a loan a lot quicker. The natural question to ask here is: why are banks asking us to pay the lower amount for much longer then? Are they trying to cheat us out of our money? There is no simple answer to this. Banks provide a service to us in return for a fee, so they are out to make a proﬁt. However, they need to be careful not to cheat their customers for fear that they’ll simply use another bank. The central issue here is one of scale. For us, the changes involved appear big. We are paying oﬀ our loan 6 years earlier by paying just a bit more a month. To a bank, however, it doesn’t matter much either way. In all likelihoxod, it doesn’t aﬀect their proﬁt margins one bit! Remember that a bank calculates repayment amount using the same methods as we’ve been learning. Therefore, they are correct amounts for given interest rates and terms. As a result, which amount is repaid does generally make a bank more or less money. It’s a simple matter of less money now or more money later. Banks generally use a 20 year repayment period by default. Learning about ﬁnancial mathematics enables you to duplicate these calculations for yourself. This way, you can decide what’s best for you. You can decide how much you want to repay each month and you’ll know of its eﬀects. A bank wouldn’t care much either way, so you should pick something that suits you.

Worked Example 165: Monthly Payments Question: Stefan and Marna want to buy a house that costs R 1 200 000. Their parents oﬀer to put down a 20% payment towards the cost of the house. They need to get a moratage for the balance. What are their monthly repayments if the term of the home loan is 30 years and the interest is 7,5%, compounded monthly ? Answer Step 1 : Determine how much money they need to borrow R1 200 00 − R240 000 = R960 000 Step 2 : Determine how to approach the problem Use the formula: P = Where P = 960 000 n = 30 × 12 = 360months i = 0,075 ÷ 12 = 0,00625 Step 3 : Solve the problem x[1 − (1 + 0,00625)−360] 0,00625 x(143,017 627 3) R6 712,46 x[1 − (1 + i)−n ] i

R960 000 = = x =

Step 4 : Write the ﬁnal answer The monthly repayments = R6 712,46

488

CHAPTER 37. FINANCE - GRADE 12

37.5

37.4.2

Exercises - Investments and Loans

1. A property costs R1 800 000. Calculate the monthly repayments if the interest rate is 14% p.a. compounded monthly and the loan must be paid of in 20 years time. 2. A loan of R 4 200 is to be returned in two equal annual instalments. If the rate of interest os 10% per annum, compounded annually, calculate the amount of each instalment.

37.4.3

Calculating Capital Outstanding

As deﬁned in Section 37.4.1, Capital outstanding is the amount we still owe the people we borrowed money from at a given moment in time. We also saw how we can calculate this using loan schedules. However, there is a signiﬁcant disadvantage to this method: it is very time consuming. For example, in order to calculate how much capital is still outstanding at time 12 using the loan schedule, we’ll have to ﬁrst calculate how much capital is outstanding at time 1 through to 11 as well. This is already quite a bit more work than we’d like to do. Can you imagine calculating the amount outstanding after 10 years (time 120)? Fortunately, there is an easier method. However, it is not immediately why this works, so let’s take some time to examine the concept. Prospective method for Capital Outstanding Let’s say that after a certain number of years, just after we made a repayment, we still owe amount Y . What do we know about Y ? We know that using the loan schedule, we can calculate what it equals to, but that is a lot of repetitive work. We also know that Y is the amount that we are still going to pay oﬀ. In other words, all the repayments we are still going to make in the future will exactly pay oﬀ Y . This is true because in the end, after all the repayments, we won’t be owing anything. Therefore, the present value of all outstanding future payments equal the present amount outstanding. This is the prospective method for calculating capital outstanding. Let’s return to a previous example. Recall the case where we were trying to repay a loan of R200 000 over 20 years. At an interested rate of 9% compounded monthly, the monthly repayment is R1 709,48. In table 37.1, we can see that after 12 month, the amount outstanding is R186 441,84. Let’s try to work this out using the the prospective method. After time 12, there is still 19 × 12 = 228 repayments left of R1 709,48 each. The present value is:

n = i = Y = =

228 0,75% R1 709,48 × R186 441,92 1 − 1,0075−228 0,0075

Oops! This seems to be almost right, but not quite. We should have got R186 441,84. We are 8 cents out. However, this is in fact not a mistake. Remember that when we worked out the monthly repayments, we rounded to the nearest cents and arrived at R1 709,48. This was because one cannot make a payment for a fraction of a cent. Therefore, the rounding oﬀ error was carried through. That’s why the two ﬁgures don’t match exactly. In ﬁnancial mathematics, this is largely unavoidable.

37.5

Formulae Sheet

As an easy reference, here are the key formulae that we derived and used during this chapter. While memorising them is nice (there are not many), it is the application that is useful. Financial 489

37.6

CHAPTER 37. FINANCE - GRADE 12

experts are not paid a salary in order to recite formulae, they are paid a salary to use the right methods to solve ﬁnancial problems.

37.5.1
P i n iT

Deﬁnitions

Principal (the amount of money at the starting point of the calculation) interest rate, normally the eﬀective rate per annum period for which the investment is made the interest rate paid T times per annum, i.e. iT = Nominal Interest Rate T

Important: Always keep the interest and the time period in the same units of time (e.g. both in years, or both in months etc.).

37.6

End of Chapter Exercises

1. Thabo is about to invest his R8 500 bonus in a special banking product which will pay 1% per annum for 1 month, then 2% per annum for the next 2 months, then 3% per annum for the next 3 months, 4% per annum for the next 4 months, and 0% for the rest of the year. The are going to charge him R100 to set up the account. How much can he expect to get back at the end of the period? 2. A special bank account pays simple interest of 8% per annum. Calculate the opening balance required to generate a closing balance of R5 000 after 2 years. 3. A diﬀerent bank account pays compound interest of 8% per annum. Calculate the opening balance required to generate a closing balance of R5 000 after 2 years. 4. Which of the two answers above is lower, and why? 5. After 7 months after an initial deposit, the value of a bank account which pays compound interest of 7,5% per annum is R3 650,81. What was the value of the initial deposit? 6. Suppose you invest R500 this year compounded at interest rate i for a year in Bank T. In the following year you invest the accumulation that you received for another year at the same interest rate and on the third year, you invested the accumulation you received at the same interest rate too. If P represents the present value (R500), ﬁnd a pattern for this investment. [Hint: ﬁnd a formula] 7. Thabani and Lungelo are both using UKZN Bank for their saving. Suppose Lungelo makes a deposit of X today at interest rate of i for six years. Thabani makes a deposit of 3X at an interest rate of 0.05. Thabani made his deposit 3 years after Lungelo made his ﬁrst deposit. If after 6 years, their investments are equal, calculate the value of i and ﬁnd X. if the sum of their investment is R20 000, use X you got to ﬁnd out how much Thabani got in 6 years. 490

CHAPTER 37. FINANCE - GRADE 12

37.6

8. Sipho invests R500 at an interest rate of log(1,12) for 5 years. Themba, Sipho’s sister invested R200 at interest rate i for 10 years on the same date that her brother made his ﬁrst deposit. If after 5 years, Themba’s accumulation equals Sipho’s, ﬁnd the interest rate i and ﬁnd out whether Themba will be able to buy her favorite cell phone after 10 years which costs R2 000. 9. Moira deposits R20 000 in her saving account for 2 years at an interest rate of 0.05. After 2 years, she invested her accumulation for another 2 years, at the same interest rate. After 4 years, she invested her accumulation for which she got for another 2 years at an interest rate of 5 %. After 6 years she choose to buy a car which costs R26 000. Her husband, Robert invested the same amount at interest rate of 5 % for 6 years. A Without using any numbers, ﬁnd a pattern for Moira’s investment? B How Moira’s investment diﬀer from Robert’s? 10. Calculate the real cost of a loan of R10 000 for 5 years at 5% capitalised monthly and half yearly. 11. Determine how long, in years, it will take for the value of a motor vehicle to decrease to 25% of its original value if the rate of depreciation, based on the reducing-balance method, is 21% per annum. 12. Andr´ and Thoko, decided to invest their winnings (amounting to R10 000) from their e science project. They decided to divide their winnings according to the following: Because Andr was the head of the project and he spent more time on it, Andr´ got 65,2 % of the e winnings and Thoko got 34,8%. So, Thoko decided to invest only 0,5 % of the share of her sum and Andr´decided to invest 1,5 % of the share of his sum. When they calculated how e much each contributed in the investment, Thoko had 25 % and Andr´ had 75 % share. e They planned to invest their money for 20 years , but, as a result of Thoko ﬁnding a job in Australia 7 years after their initial investment. They both decided to take whatever value was there and split it according to their initial investment(in terms of percentages). Find how much each will get after 7 years, if the interest rate is equal to the percentage that Thoko invested (NOT the money but the percentage).

491

37.6

CHAPTER 37. FINANCE - GRADE 12

492

Chapter 38

Factorising Cubic Polynomials Grade 12
38.1 Introduction

In grades 10 and 11, you learnt how to solve diﬀerent types of equations. Most of the solutions, relied on being able to factorise some expression and the factorisation of quadratics was studied in detail. This chapter focusses on the factorisation of cubic polynomials, that is expressions with the highest power equal to 3.

38.2

The Factor Theorem

The factor theorem describes the relationship between the root of a polynomial and a factor of the polynomial. Deﬁnition: Factor Theorem For any polynomial, f (x), for all values of a which satisfy f (a) = 0, (x − a) is a factor of f (x). Or, more concisely: f (x) = q(x) x−a is a polynomial. In other words: If the remainder when dividing f (x) by (x − a) is zero, then (ax + b) is a factor of f (x). b So if f (− a ) = 0, then (ax + b) is a factor of f (x).

Cubic expressions have a highest power of 3 on the unknown variable. This means that there should be at least 3 factors. We have seen in Grade 10 that the sum and diﬀerence of cubes is factorised as follows.: (x + y)(x2 − xy + y 2 ) = x3 + y 3 and (x − y)(x2 + xy + y 2 ) = x3 − y 3 We also saw that the quadratic terms do not have rational roots. There are many methods of factorising a cubic polynomial. The general method is similar to that used to factorise quadratic equations. If you have a cubic polynomial of the form: f (x) = ax3 + bx2 + cx + d then you should expect factors of the form: (Ax + B)(Cx + D)(Ex + F ). (38.1)

We will deal with simplest case ﬁrst. When a = 1, then A = C = E = 1, and you only have to determine B, D and F . For example, ﬁnd the factors of: x3 − 2x2 − 5x + 6. In this case we have a b = 1 = −2

This is a set of three equations in three unknowns. However, we know that B, D and F are factors of 6 because BDF = 6. Therefore we can use a trial and error method to ﬁnd B, D and F . This can become a very tedious method, therefore the Factor Theorem can be used to ﬁnd the factors of cubic polynomials.

In general, to factorise a cubic polynomial, you ﬁnd one factor by trial and error. Use the factor theorem to conﬁrm that the guess is a root. Then divide the cubic polynomial by the factor to obtain a quadratic. Once you have the quadratic, you can apply the standard methods to factorise the quadratic. For example the factors of x3 − 2x2 − 5x + 6 can be found as follows: There are three factors which we can write as (x − a)(x − b)(x − c). 495

In grades 10 and 11 you have learnt about linear functions and quadratic functions as well as the hyperbolic functions and exponential functions and many more. In grade 12 you are expected to demonstrate the ability to work with various types of functions and relations including the inverses of some functions and generate graphs of the inverse relations of functions, in particular the inverses of: y = ax + q y = ax2 y = ax; a > 0 .

39.2

Deﬁnition of a Function

A function is a relation for which there is only one value of y corresponding to any value of x. We sometimes write y = f (x), which is notation meaning ’y is a function of x’. This deﬁnition makes complete sense when compared to our real world examples — each person has only one height, so height is a function of people; on each day, in a speciﬁc town, there is only one average temperature. However, some very common mathematical constructions are not functions. For example, consider the relation x2 + y 2 = 4. This relation describes a circle of radius 2 centred at the origin, as in ﬁgure 39.1. If we let x = 0, we see that y 2 = 4 and thus either y = 2 or y = −2. Since there are two y values which are possible for the same x value, the relation x2 + y 2 = 4 is not a function. There is a simple test to check if a relation is a function, by looking at its graph. This test is called the vertical line test. If it is possible to draw any vertical line (a line of constant x) which crosses the relation more than once, then the relation is not a function. If more than one intersection point exists, then the intersections correspond to multiple values of y for a single value of x. We can see this with our previous example of the circle by looking at its graph again in Figure 39.1. We see that we can draw a vertical line, for example the dotted line in the drawing, which cuts the circle more than once. Therefore this is not a function.

39.2.1

Exercises

1. State whether each of the following equations are functions or not: A x+y =4 501

39.3

CHAPTER 39. FUNCTIONS AND GRAPHS - GRADE 12

2 1

−2

−1

1 −1 −2

2

Figure 39.1: Graph of y 2 + x2 = 4

B y=x 4 C y = 2x D x2 + y 2 = 4 2. The table gives the average per capita income, d, in a region of the country as a function of the percent unemployed, u. Write down the equation to show that income is a function of the persent unemployed. u d 1 22500 2 22000 3 21500 4 21000

39.3

Notation used for Functions

In grade 10 you were introduced to the notation used to ”name” a function. In a function y = f (x), y is called the dependent variable, because the value of y depends on what you choose as x. We say x is the independent variable, since we can choose x to be any number. Similarly if g(t) = 2t + 1, then t is the independent variable and g is the function name. If f (x) = 3x − 5 and you are ask to determine f (3), then you have to work out the value for f (x) when x = 3. For example, f (x) = f (3) = = 3x − 5 3(3) − 5

4

39.4

Graphs of Inverse Functions

In earlier grades, you studied various types of functions and understood the eﬀect of various parameters in the general equation. In this section, we will consider inverse functions. An inverse function is a function which ”does the reverse” of a given function. More formally, if f is a function with domain X, then f −1 is its inverse function if and only if for every x ∈ X we have: f −1 (f (x)) = f (f −1 (x)) = x (39.1) For example, if the function x → 3x + 2 is given, then its inverse function is x → is usually written as: f f −1 : : x → 3x + 2 (x − 2) x→ 3 502 (x − 2) . This 3 (39.2) (39.3)

CHAPTER 39. FUNCTIONS AND GRAPHS - GRADE 12 The superscript ”-1” is not an exponent. If a function f has an inverse then f is said to be invertible.

39.4

If f is a real-valued function, then for f to have a valid inverse, it must pass the horizontal line test, that is a horizontal line y = k placed anywhere on the graph of f must pass through f exactly once for all real k. It is possible to work around this condition, by deﬁning a multi-valued function as an inverse. If one represents the function f graphically in a xy-coordinate system, then the graph of f −1 is the reﬂection of the graph of f across the line y = x. Algebraically, one computes the inverse function of f by solving the equation y = f (x) for x, and then exchanging y and x to get y = f −1 (x)

The inverse function of a straight line is also a straight line. For example, the straight line equation given by y = 2x − 3 has as inverse the function, y = 1 3 2 x + 2 . The graphs of these functions are shown in Figure 39.2. It can be seen that the two graphs are reﬂections of each other across the line y = x.

We have seen that the domain of a function of the form y = ax + q is {x : x ∈ R} and the range is {y : y ∈ R}. Since the inverse function of a straight line is also a straight line, the inverse function will have the same domain and range as the original function.

Intercepts
q 1 The general form of the inverse function of the form y = ax + q is y = a x − a .

q By setting x = 0 we have that the y-intercept is yint = − a . Similarly, by setting y = 0 we have that the x-intercept is xint = q. q 1 It is interesting to note that if f (x) = ax + q, then f −1 (x) = a x − a and the y-intercept of −1 f (x) is the x-intercept of f (x) and the x-intercept of f (x) is the y-intercept of f −1 (x).

39.4.2

Exercises

1. Given f (x) = 2x − 3, ﬁnd f −1 (x) 2. Consider the function f (x) = 3x − 7. A Is the relation a function? B Identify the domain and range. 3. Sketch the graph of the function f (x) = 3x − 1 and its inverse on the same set of axes. 4. The inverse of a function is f −1 (x) = 2x − 4, what is the function f (x)?

We see that the inverse function of y = ax2 is not a function because it fails the vertical line √ test. If we draw a vertical line through the graph of f −1 (x) = ± x, the line intersects the graph more than once. There has to be a restriction on the domain of a parabola for the inverse to also be a function. Consider the function f (x) = −x2 + 9. The inverse of f can be found by witing f (y) = x. Then x = y2 = y = −y 2 + 9 9−x √ ± 9−x

√ √ If x ≥ 0, then 9 − x is a function. If the restriction on the domain of f is x ≤ 0 then − 9 − x would be a function.

39.4.4

Exercises

1. The graph of f −1 is shown. Find the equation of f , given that the graph of f is a parabola. (Do not simplify your answer) 504

2. f (x) = 2x2 . A Draw the graph of f and state its domain and range. B Find f −1 and state the domain and range. C What must the domain of f be, so that f −1 is a function ? 3. Sketch the graph of x = − 10 − y 2 . Label a point on the graph other than the intercepts with the axes. 4. A Sketch the graph of y = x2 labelling a point other than the origin on your graph. B Find the equation of the inverse of the above graph in the form y = . . .. √ C Now sketch the y = x. √ D The tangent to the graph of y = x at the point A(9;3) intersects the x-axis at B. Find the equation of this tangent and hence or otherwise prove that the y-axis bisects the straight line AB. 5. Given: g(x) = −1 + √ x, ﬁnd the inverse of g(x) in the form g −1 (x). 505

Figure 39.4: The function f (x) = 10x and its inverse f −1 (x) = log(x). The line y = x is shown as a dashed line. The exponential function and the logarithmic function are inverses of each other; the graph of the one is the graph of the other, reﬂected in the line y = x. The domain of the function is equal to the range of the inverse. The range of the function is equal to the domain of the inverse.

39.4.6

Exercises

1 1. Given that f (x) = [ 5 ]x , sketch the graphs of f and f −1 on the same system of axes indicating a point on each graph (other than the intercepts) and showing clearly which is f and which is f −1 .

2. Given that f (x) = 4−x , A Sketch the graphs of f and f −1 on the same system of axes indicating a point on each graph (other than the intercepts) and showing clearly which is f and which is f −1 . B Write f −1 in the form y = . . .. √ 3. Given g(x) = −1 + x, ﬁnd the inverse of g(x) in the form g −1 (x) = . . . 4. A B C D

Sketch the graph of y = x2 , labeling a point other than the origin on your graph. Find the equation of the inverse of the above graph in the form y = . . . √ Now, sketch y = x. √ The tangent to the graph of y = x at the point A(9; 3) intersects the x-axis at B. Find the equation of this tangent, and hence, or otherwise, prove that the y-axis bisects the straight line AB. 506

6. Given the equation h(x) = 3x A Write down the inverse in the form h−1 (x) = ... B Sketch the graphs of h(x) and h−1 (x) on teh same set of axes, labelling the intercepts with the axes. C For which values of x is h−1 (x) undeﬁned ? 7. A Sketch the graph of y = x2 , labelling a point other than the origin on your graph. B Find the equation of the inverse of the above graph in the form y = . . . √ C Now, sketch y = x. √ D The tangent to the graph of y = x at the point A(9; 3) intersects the x-axis at B. Find the equation of this tangent, and hence, or otherwise, prove that the y-axis bisects the straight line AB.

508

Chapter 40

Diﬀerential Calculus - Grade 12
40.1 Why do I have to learn this stuﬀ?

Calculus is one of the central branches of mathematics and was developed from algebra and geometry. Calculus is built on the concept of limits, which will be discussed in this chapter. Calculus consists of two complementary ideas: diﬀerential calculus and integral calculus. Only diﬀerential calculus will be studied. Diﬀerential calculus is concerned with the instantaneous rate of change of quantities with respect to other quantities, or more precisely, the local behaviour of functions. This can be illustrated by the slope of a function’s graph. Examples of typical diﬀerential calculus problems include: ﬁnding the acceleration and velocity of a free-falling body at a particular moment and ﬁnding the optimal number of units a company should produce to maximize its proﬁt. Calculus is fundamentally diﬀerent from the mathematics that you have studied previously. Calculus is more dynamic and less static. It is concerned with change and motion. It deals with quantities that approach other quantities. For that reason it may be useful to have an overview of the subject before beginning its intensive study. Calculus is a tool to understand many natural phenomena like how the wind blows, how water ﬂows, how light travels, how sound travels and how the planets move. However, other human activities such as economics are also made easier with calculus. In this section we give a glimpse of some of the main ideas of calculus by showing how limits arise when we attempt to solve a variety of problems. Extension: Integral Calculus Integral calculus is concerned with the accumulation of quantities, such as areas under a curve, linear distance traveled, or volume displaced. Diﬀerential and integral calculus act inversely to each other. Examples of typical integral calculus problems include ﬁnding areas and volumes, ﬁnding the amount of water pumped by a pump with a set power input but varying conditions of pumping losses and pressure and ﬁnding the amount of rain that fell in a certain area if the rain fell at a speciﬁc rate.

teresting Both Isaac Newton (4 January 1643 – 31 March 1727) and Gottfried Liebnitz Interesting Fact Fact (1 July 1646 – 14 November 1716 (Hanover, Germany)) are credited with the ‘invention’ of calculus. Newton was the ﬁrst to apply calculus to general physics, while Liebnitz developed most of the notation that is still in use today.
When Newton and Leibniz ﬁrst published their results, there was some controversy over whether Leibniz’s work was independent of Newton’s. While Newton derived his results years before Leibniz, it was only some time after Leibniz published in 1684 that Newton published. Later, Newton would claim that Leibniz 509

40.2

CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12 got the idea from Newton’s notes on the subject; however examination of the papers of Leibniz and Newton show they arrived at their results independently, with Leibniz starting ﬁrst with integration and Newton with diﬀerentiation. This controversy between Leibniz and Newton divided English-speaking mathematicians from those in Europe for many years, which slowed the development of mathematical analysis. Today, both Newton and Leibniz are given credit for independently developing calculus. It is Leibniz, however, who is credited with giving the new discipline the name it is known by today: ”calculus”. Newton’s name for it was ”the science of ﬂuxions”.

40.2
40.2.1

Limits
A Tale of Achilles and the Tortoise

teresting Zeno (circa 490 BC - circa 430 BC) was a pre-Socratic Greek philosopher of Interesting Fact Fact southern Italy who is famous for his paradoxes.

One of Zeno’s paradoxes can be summarised by:

Achilles and a tortoise agree to a race, but the tortoise is unhappy because Achilles is very fast. So, the tortoise asks Achilles for a head-start. Achilles agrees to give the tortoise a 1 000 m head start. Does Achilles overtake the tortoise?

However, Zeno (the Greek philosopher who thought up this problem) looked at it as follows: Achilles takes 1000 = 500 s t= 2 to travel the 1 000 m head start that the tortoise had. However, in this 500 s, the tortoise has travelled a further x = (500)(0,25) = 125 m. 125 = 62,5 s 2 to travel the 125 m. In this 62,5 s, the tortoise travels a further t= x = (62,5)(0,25) = 15,625 m. Zeno saw that Achilles would always get closer but wouldn’t actually overtake the tortoise. Achilles then takes another

40.2.2

Sequences, Series and Functions

So what does Zeno, Achilles and the tortoise have to do with calculus? Well, in Grades 10 and 11 you studied sequences. For the sequence 1 2 3 4 0, , , , , . . . 2 3 4 5 which is deﬁned by the expression 1 n the terms get closer to 1 as n gets larger. Similarly, for the sequence an = 1 − 1 1 1 1 1, , , , , . . . 2 3 4 5 which is deﬁned by the expression 1 n the terms get closer to 0 as n gets larger. We have also seen that the inﬁnite geometric series has a ﬁnite total. The inﬁnite geometric series is an = S∞ =
∞ i=1

a1 .ri−1 =

a1 1−r

for

−1

where a1 is the ﬁrst term of the series and r is the common ratio. 511

40.2

CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12

We see that there are some functions where the value of the function gets close to or approaches a certain value. x2 + 4x − 12 x+6 The numerator of the function can be factorised as: y= y= (x + 6)(x − 2) . x+6 Similarly, for the function:

Then we can cancel the x − 6 from numerator and denominator and we are left with: y = x − 2. However, we are only able to cancel the x + 6 term if x = −6. If x = −6, then the denominator becomes 0 and the function is not deﬁned. This means that the domain of the function does not include x = −6. But we can examine what happens to the values for y as x gets close to -6. These values are listed in Table 40.1 which shows that as x gets closer to -6, y gets close to 8. (x + 6)(x − 2) as x gets close to -6. x+6

The graph of this function is shown in Figure 40.1. The graph is a straight line with slope 1 and intercept -2, but with a missing section at x = −6. Extension: Continuity We say that a function is continuous if there are no values of the independent variable for which the function is undeﬁned.

Worked Example 172: Limits Notation Question: Summarise the following situation by using limit notation: As x gets close to 1, the value of the function y =x+2 gets close to 3. 513

40.2 Answer This is written as:

CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12

x→1

lim x + 2 = 3

in limit notation.

We can also have the situation where a function has a diﬀerent value depending on whether x approaches from the left or the right. An example of this is shown in Figure 40.2.

4 3 2 1 −7 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 5 6 7

1 Figure 40.2: Graph of y = x .

1 As x → 0 from the left, y = x approaches −∞. As x → 0 from the right, y = +∞. This is written in limits notation as:

1 x

approaches

x→0−

lim

1 = −∞ x

for x approaching zero from the left and lim 1 =∞ x

x→0+

for x approaching zero from the right. You can calculate the limit of many diﬀerent functions using a set method. Method: Limits If you are required to calculate a limit like limx→a then: 1. Simplify the expression completely. 2. If it is possible, cancel all common terms. 3. Let x approach the a.

In Grade 10 you learnt about average gradients on a curve. The average gradient between any two points on a curve is given by the gradient of the straight line that passes through both points. In Grade 11 you were introduced to the idea of a gradient at a single point on a curve. We saw that this was the gradient of the tangent to the curve at the given point, but we did not learn how to determine the gradient of the tangent. Now let us consider the problem of trying to ﬁnd the gradient of a tangent t to a curve with equation y = f (x) at a given point P . tangent P f (x) We know how to calculate the average gradient between two points on a curve, but we need two points. The problem now is that we only have one point, namely P . To get around the problem we ﬁrst consider a secant to the curve that passes through point P and another point on the curve Q. We can now ﬁnd the average gradient of the curve between points P and Q. secant f (a) f (a − h) P Q f (x)

a−h

a

If the x-coordinate of P is a, then the y-coordinate is f (a). Similarly, if the x-coordinate of Q is a − h, then the y-coordinate is f (a − h). If we choose a as x2 and a − h as x1 , then: y1 = f (a − h) y2 = f (a). We can now calculate the average gradient as: y2 − y1 x2 − x1 = = f (a) − f (a − h) a − (a − h) f (a) − f (a − h) h (40.12) (40.13)

Now imagine that Q moves along the curve toward P . The secant line approaches the tangent line as its limiting position. This means that the average gradient of the secant approaches the gradient of the tangent to the curve at P . In (40.13) we see that as point Q approaches point P , h gets closer to 0. When h = 0, points P and Q are equal. We can now use our knowledge of limits to write this as: gradient at P = lim
h→0

f (a) − f (a − h) . h

(40.14)

and we say that the gradient at point P is the limit of the average gradient as Q approaches P along the curve. 516

CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12

40.2

Activity :: Investigation : Limits The gradient at a point x on a curve deﬁned by f (x) can also be written as: lim f (x + h) − f (x) h (40.15)

h→0

Show that this is equivalent to (40.14).

Worked Example 176: Limits Question: For the function f (x) = 2x2 − 5x, determine the gradient of the tangent to the curve at the point x = 2. Answer Step 1 : Calculating the gradient at a point We know that the gradient at a point x is given by: lim f (x + h) − f (x) h

The tangent problem has given rise to the branch of calculus called diﬀerential calculus and the equation: f (x + h) − f (x) lim h→0 h deﬁnes the derivative of the function f (x). Using (40.15) to calculate the derivative is called ﬁnding the derivative from ﬁrst principles.

There are a few diﬀerent notations used to refer to derivatives. If we use the traditional notation y = f (x) to indicate that the dependent variable is y and the independent variable is x, then some common alternative notations for the derivative are as follows: f ′ (x) = y ′ = dy df d = = f (x) = Df (x) = Dx f (x) dx dx dx

d The symbols D and dx are called diﬀerential operators because they indicate the operation of diﬀerentiation, which is the process of calculating a derivative. It is very important that you learn to identify these diﬀerent ways of denoting the derivative, and that you are consistent in your usage of them when answering questions.

dy Important: Though we choose to use a fractional form of representation, dx is a limit and dy dy is not a fraction, i.e. dx does not mean dy ÷ dx. dx means y diﬀerentiated with respect to dp d x. Thus, dx means p diﬀerentiated with respect to x. The ‘ dx ’ is the “operator”, operating on some function of x.

Worked Example 178: Derivatives - First Principles Question: Calculate the derivative of g(x) = x − 1 from ﬁrst principles. Answer Step 1 : Calculating the gradient at a point We know that the gradient at a point x is given by: g ′ (x) = lim g(x + h) − g(x) h→0 h 519

Thus far we have learnt about how to diﬀerentiate various functions, but I am sure that you are beginning to ask, What is the point of learning about derivatives? Well, we know one important fact about a derivative: it is a gradient. So, any problems involving the calculations of gradients or rates of change can use derivatives. One simple application is to draw graphs of functions by ﬁrstly determine the gradients of straight lines and secondly to determine the turning points of the graph.

40.5.1

Finding Equations of Tangents to Curves

In section 40.2.4 we saw that ﬁnding the gradient of a tangent to a curve is the same as ﬁnding the slope of the same curve at the point of the tangent. We also saw that the gradient of a function at a point is just its derivative. Since we have the gradient of the tangent and the point on the curve through which the tangent passes, we can ﬁnd the equation of the tangent.

Worked Example 181: Finding the Equation of a Tangent to a Curve Question: Find the equation of the tangent to the curve y = x2 at the point (1,1) and draw both functions. Answer Step 1 : Determine what is required We are required to determine the equation of the tangent to the curve deﬁned by y = x2 at the point (1,1). The tangent is a straight line and we can ﬁnd the equation by using derivatives to ﬁnd the gradient of the straight line. Then we will have the gradient and one point on the line, so we can ﬁnd the equation using: y − y1 = m(x − x1 ) from grade 11 Coordinate Geometry. Step 2 : Diﬀerentiate the function Using our rules of diﬀerentiation we get: y ′ = 2x Step 3 : Find the gradient at the point (1,1) In order to determine the gradient at the point (1,1), we substitute the x-value into the equation for the derivative. So, y ′ at x = 1 is: 2(1) = 2 Step 4 : Find equation of tangent y − y1 y−1 y y = = m(x − x1 ) (2)(x − 1)

Diﬀerentiation can be used to sketch the graphs of functions, by helping determine the turning points. We know that if a graph is increasing on an interval and reaches a turning point, then the graph will start decreasing after the turning point. The turning point is also known as a stationary point because the gradient at a turning point is 0. We can then use this information to calculate turning points, by calculating the points at which the derivative of a function is 0. Important: If x = a is a turning point of f (x), then: f ′ (a) = 0 This means that the derivative is 0 at a turning point. Take the graph of y = x2 as an example. We know that the graph of this function has a turning point at (0,0), but we can use the derivative of the function: y ′ = 2x and set it equal to 0 to ﬁnd the x-value for which the graph has a turning point. 2x = 0 x = 0 We then substitute this into the equation of the graph (i.e. y = x2 ) to determine the y-coordinate of the turning point: f (0) = (0)2 = 0 This corresponds to the point that we have previously calculated. 524

We are now ready to sketch graphs of functions. Method: Sketching GraphsSuppose we are given that f (x) = ax3 + bx2 + cx + d, then there are ﬁve steps to be followed to sketch the graph of the function:

1. If a > 0, then the graph is increasing from left to right, and has a maximum and then a minimum. As x increases, so does f (x). If a < 0, then the graph decreasing is from left to right, and has ﬁrst a minimum and then a maximum. f (x) decreases as x increases. 2. Determine the value of the y-intercept by substituting x = 0 into f (x) 3. Determine the x-intercepts by factorising ax3 + bx2 + cx + d = 0 and solving for x. First try to eliminate constant common factors, and to group like terms together so that the expression is expressed as economically as possible. Use the factor theorem if necessary. 525

40.5

CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12
df dx

4. Find the turning points of the function by working out the derivative zero, and solving for x.

and setting it to

5. Determine the y-coordinates of the turning points by substituting the x values obtained in the previous step, into the expression for f (x). 6. Draw a neat sketch.

which does not have real roots. Therefore, the graph of g(x) does not have any x-intercepts. Step 3 : Find the turning points of the function dg Work out the derivative dx and set it to zero to for the x coordinate of the turning point. dg = 2x − 1 dx dg = dx 2x − 1 = 2x = x =

0 0 1 1 2

Step 4 : Determine the y-coordinates of the turning points by substituting the x values obtained in the previous step, into the expression for f (x). 1 y coordinate of turning point is given by calculating g( 2 ). 1 g( ) = 2 = =
7 The turning point is at ( 1 , 4 ) 2 Step 5 : Draw a neat sketch

Exercise: Sketching Graphs 1. Given f (x) = x3 + x2 − 5x + 3: A Show that (x − 1) is a factor of f (x) and hence fatorise f (x) fully. B Find the coordinates of the intercepts with the axes and the turning points and sketch the graph

dy If the derivative ( dx ) is zero at a point, the gradient of the tangent at that point is zero. It means that a turning point occurs as seen in the previous example.

y 9 8 7 6 5 (3;4) 4 3 2 1 −1 −1 (1;0) 1 2 3 4 (4;0) x

From the drawing the point (1;0) represents a local minimum and the point (3;4) the local maximum. A graph has a horizontal point of inﬂexion where the derivative is zero but the sign of the sign of the gradient does not change. That means the graph always increases or always decreases. 529

40.6 y

CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12

(3;1) x

From this drawing, the point (3;1) is a horizontal point of inﬂexion, because the sign of the derivative stays positive.

40.6

Using Diﬀerential Calculus to Solve Problems

We have seen that diﬀerential calculus can be used to determine the stationary points of functions, in order to sketch their graphs. However, determining stationary points also lends itself to the solution of problems that require some variable to be optimised. For example, if fuel used by a car is deﬁned by: f (v) = 3 2 v − 6v + 245 80 (40.20)

where v is the travelling speed, what is the most economical speed (that means the speed that uses the least fuel)? If we draw the graph of this function we ﬁnd that the graph has a minimum. The speed at the minimum would then give the most economical speed.

We have seen that the coordinates of the turning point can be calculated by diﬀerentiating the function and ﬁnding the x-coordinate (speed in the case of the example) for which the derivative is 0. Diﬀerentiating (40.20), we get: 3 v−6 40 If we set f ′ (v) = 0 we can calculate the speed that corresponds to the turning point. 530 f ′ (v) =

CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12

40.6

f ′ (v) 0 v

3 v−6 40 3 v−6 = 40 6 × 40 = 3 = 80 =

This means that the most economical speed is 80 km·hr−1 .

Worked Example 185: Optimisation Problems Question: The sum of two positive numbers is 10. One of the numbers is multiplied by the square of the other. If each number is greater than 0, ﬁnd the numbers that make this product a maximum. Answer Step 1 : Examine the problem and formulate the equations that are required Let the two numbers be a and b. Then we have: a + b = 10 (40.21)

Step 4 : Write the ﬁnal answer The product is maximised if a and b are both equal to 5.

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CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12

Worked Example 186: Optimisation Problems Question: Michael wants to start a vegetable garden, which he decides to fence oﬀ in the shape of a rectangle from the rest of the garden. Michael only has 160 m of fencing, so he decides to use a wall as one border of the vegetable garden. Calculate the width and length of the garden that corresponds to largest possible area that Michael can fence oﬀ. wall length, l

garden

width, w Answer Step 1 : Examine the problem and formulate the equations that are required The important pieces of information given are related to the area and modiﬁed perimeter of the garden. We know that the area of the garden is: A= w·l (40.25)

We are also told that the fence covers only 3 sides and the three sides should add up to 160 m. This can be written as: 160 = w + l + l However, we can use (40.26) to write w in terms of l: w = 160 − 2l Substitute (40.27) into (40.25) to get: A = (160 − 2l)l = 160l − 2l2 (40.28) (40.27) (40.26)

Exercise: Solving Optimisation Problems using Diﬀerential Calculus 1. The sum of two positive numbers is 20. One of the numbers is multiplied by the square of the other. Find the numbers that make this products a maximum. 2. A wooden block is made as shown in the diagram. The ends are right-angled triangles having sides 3x, 4x and 5x. The length of the block is y. The total surface area of the block is 3 600 cm2 .

3x

4x

y

300 − x2 . x B Find the value of x for which the block will have a maximum volume. (Volume = area of base × height.) A Show that y = 3. The diagram shows the plan for a verandah which is to be built on the corner of a cottage. A railing ABCDE is to be constructed around the four edges of the verandah. y C D

x

verandah F B A E cottage If AB = DE = x and BC = CD = y, and the length of the railing must be 30 metres, ﬁnd the values of x and y for which the verandah will have a maximum area.

533

40.6

CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12

40.6.1

Rate of Change problems
f (b)−f (a) b−a

Two concepts were discussed in this chapter: Average rate of change = limh→0 f (x+h)−f (x) . h

and Instan-

taneous rate of change = When we mention rate of change, the latter is implied. Instantaneous rate of change is the derivative. When Average rate of change is required, it will be speciﬁcally refer to as average rate of change. Velocity is one of the most common forms of rate of change. Again, average velocity = average rate of change and instantaneous velocity = instantaneous rate of change = derivative. Velocity refers to the increase of distance(s) for a corresponding increade in time (t). The notation commonly used for this is: v(t) = ds = s′ (t) dt

Acceleration is the change in velocity for a corersponding increase in time. Therefore, acceleration is the derivative of velocity a(t) = v ′ (t) This implies that acceleration is the second derivative of the distance(s).

Worked Example 187: Rate of Change Question: The height (in metres) of a golf ball that is hit into the air after t seconds, is given by h(t) = 20t = 5t2 . Determine 1. the average velocity of the ball during the ﬁrst two seconds 2. the velocity of the ball after 1,5 seconds 3. when the velocity is zero 4. the velocity at which the ball hits the ground 5. the acceleration of the ball Answer Step 1 : Average velocity h(2) − h(0) 2−0 [20(2) − 5(2)2 ] − [20(0) − 5(0)2 ] 2 40 − 20 2 10 ms−1

A Determine the co-ordinates of the turning points of f . B Draw a neat sketch graph of f . Clearly indicate the co-ordinates of the intercepts with the axes, as well as the co-ordinates of the turning points. C For which values of k will the equation f (x) = k , have exactly two real roots? D Determine the equation of the tangent to the graph of f (x) = 2x3 − 5x2 − 4x + 3 at the point where x = 1. 6. A Sketch the graph of f (x) = x3 − 9x2 + 24x − 20, showing all intercepts with the axes and turning points. B Find the equation of the tangent to f (x) at x = 4. 7. Calculate: 1 − x3 x→1 1 − x lim f (x) = 2x2 − x A Use the deﬁnition of the derivative to calculate f ′ (x). B Hence, calculate the co-ordinates of the point at which the gradient of the tangent to the graph of f is 7. √ 9. If xy − 5 = x3 , determine dx dy 10. Given: g(x) = (x−2 + x2 )2 . Calculate g ′ (2). 11. Given: A Find: B Solve: f (x) = 2x − 3 f −1 (x) f −1 (x) = 3f ′ (x)

8. Given:

12. Find f ′ (x) for each of the following: √ 5 x3 + 10 A f (x) = 3 (2x2 − 5)(3x + 2) B f (x) = x2 13. Determine the minimum value of the sum of a positive number and its reciprocal. 14. If the displacement s (in metres) of a particle at time t (in seconds) is governed by the 1 equation s = 2 t3 − 2t, ﬁnd its acceleration after 2 seconds. (Acceleration is the rate of change of velocity, and velocity is the rate of change of displacement.) 15. A After doing some research, a transport company has determined that the rate at which petrol is consumed by one of its large carriers, travelling at an average speed of x km per hour, is given by: P (x) = 55 x + 2x 200 litres per kilometre

i. Assume that the petrol costs R4,00 per litre and the driver earns R18,00 per hour (travelling time). Now deduce that the total cost, C, in Rands, for a 2 000 km trip is given by: 256000 + 40x C(x) = x ii. Hence determine the average speed to be maintained to eﬀect a minimum cost for a 2 000 km trip. 536

i. Determine an expression for the rate of change of temperature with time. ii. During which time interval was the temperature dropping? 16. The depth, d, of water in a kettle t minutes after it starts to boil, is given by d = 1 1 86 − 8 t − 4 t3 , where d is measured in millimetres. A How many millimetres of water are there in the kettle just before it starts to boil? B As the water boils, the level in the kettle drops. Find the rate at which the water level is decreasing when t = 2 minutes. C How many minutes after the kettle starts boiling will the water level be dropping at 1 a rate of 12 8 mm/minute?

537

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CHAPTER 40. DIFFERENTIAL CALCULUS - GRADE 12

538

Chapter 41

Linear Programming - Grade 12
41.1 Introduction

In Grade 11 you were introduced to linear programming and solved problems by looking at points on the edges of the feasible region. In Grade 12 you will look at how to solve linear programming problems in a more general manner.

41.2

Terminology

Here is a recap of some of the important concepts in linear programming.

41.2.1

Feasible Region and Points

Constraints mean that we cannot just take any x and y when looking for the x and y that optimise our objective function. If we think of the variables x and y as a point (x,y) in the xyplane then we call the set of all points in the xy-plane that satisfy our constraints the feasible region. Any point in the feasible region is called a feasible point. For example, the constraints x≥0 y≥0 mean that every (x,y) we can consider must lie in the ﬁrst quadrant of the xy plane. The constraint x≥y means that every (x,y) must lie on or below the line y = x and the constraint x ≤ 20 means that x must lie on or to the left of the line x = 20. We can use these constraints to draw the feasible region as shown by the shaded region in Figure 41.1.

Important: ax + by = c If b = 0, feasible points must lie on the line c y = −ax + b b If b = 0, feasible points must lie on the line x = c/a If b = 0, feasible points must lie on or below the line y = − a x + c . b b If b = 0, feasible points must lie on or to the left of the line x = c/a.

ax + by ≤ c

When a constraint is linear, it means that it requires that any feasible point (x,y) lies on one side of or on a line. Interpreting constraints as graphs in the xy plane is very important since it allows us to construct the feasible region such as in Figure 41.1.

41.3

Linear Programming and the Feasible Region

If the objective function and all of the constraints are linear then we call the problem of optimising the objective function subject to these constraints a linear program. All optimisation problems we will look at will be linear programs. The major consequence of the constraints being linear is that the feasible region is always a polygon. This is evident since the constraints that deﬁne the feasible region all contribute a line segment to its boundary (see Figure 41.1). It is also always true that the feasible region is a convex polygon. The objective function being linear means that the feasible point(s) that gives the solution of a linear program always lies on one of the vertices of the feasible region. This is very important since, as we will soon see, it gives us a way of solving linear programs. We will now see why the solutions of a linear program always lie on the vertices of the feasible region. Firstly, note that if we think of f (x,y) as lying on the z axis, then the function f (x,y) = ax + by (where a and b are real numbers) is the deﬁnition of a plane. If we solve for y in the equation deﬁning the objective function then

=

f (x,y) −a x+ b b

x

(41.1)

What this means is that if we ﬁnd all the points where f (x,y) = c for any real number c (i.e. f (x,y) is constant with a value of c), then we have the equation of a line. This line we call a level line of the objective function. 540

have also been drawn in. It is very important to realise that these are not the only level lines; in fact, there are inﬁnitely many of them and they are all parallel to each other. Remember that if we look at any one level line f (x,y) has the same value for every point (x,y) that lies on that line. Also, f (x,y) will always have diﬀerent values on diﬀerent level lines. y 20 15 10 5 5 10 15 20 f (x,y) = −20 f (x,y) = −10 f (x,y) = 0 f (x,y) = 10 f (x,y) = 20 x

If a ruler is placed on the level line corresponding to f (x,y) = −20 in Figure 41.2 and moved down the page parallel to this line then it is clear that the ruler will be moving over level lines which correspond to larger values of f (x,y). So if we wanted to maximise f (x,y) then we simply move the ruler down the page until we reach the “lowest” point in the feasible region. This point will then be the feasible point that maximises f (x,y). Similarly, if we wanted to minimise f (x,y) then the “highest” feasible point will give the minimum value of f (x,y). Since our feasible region is a polygon, these points will always lie on vertices in the feasible region. The fact that the value of our objective function along the line of the ruler increases as we move it down and decreases as we move it up depends on this particular example. Some other examples might have that the function increases as we move the ruler up and decreases as we move it down. It is a general property, though, of linear objective functions that they will consistently increase or decrease as we move the ruler up or down. Knowing which direction to move the ruler in order to maximise/minimise f (x,y) = ax + by is as simple as looking at the sign of b (i.e. “is b negative, positive or zero?”). If b is positive, then f (x,y) increases as we move the ruler up and f (x,y) decreases as we move the ruler down. The opposite happens for the case when b is negative: f (x,y) decreases as we move the ruler up and f (x,y) increases as we move the ruler down. If b = 0 then we need to look at the sign of a. 541

41.3

CHAPTER 41. LINEAR PROGRAMMING - GRADE 12

If a is positive then f (x,y) increases as we move the ruler to the right and decreases if we move the ruler to the left. Once again, the opposite happens for a negative. If we look again at the objective function mentioned earlier, f (x,y) = x − 2y with a = 1 and b = −2, then we should ﬁnd that f (x,y) increases as we move the ruler down the page since b = −2 < 0. This is exactly what we found happening in Figure 41.2. The main points about linear programming we have encountered so far are • The feasible region is always a polygon. • Solutions occur at vertices of the feasible region. • Moving a ruler parallel to the level lines of the objective function up/down to the top/bottom of the feasible region shows us which of the vertices is the solution. • The direction in which to move the ruler is determined by the sign of b and also possibly by the sign of a. These points are suﬃcient to determine a method for solving any linear program. Method: Linear Programming If we wish to maximise the objective function f (x,y) then: 1. Find the gradient of the level lines of f (x,y) (this is always going to be − a as we saw in b Equation ??) 2. Place your ruler on the xy plane, making a line with gradient − a (i.e. b units on the b x-axis and −a units on the y-axis) 3. The solution of the linear program is given by appropriately moving the ruler. Firstly we need to check whether b is negative, positive or zero. A If b > 0, move the ruler up the page, keeping the ruler parallel to the level lines all the time, until it touches the “highest” point in the feasible region. This point is then the solution. B If b < 0, move the ruler in the opposite direction to get the solution at the “lowest” point in the feasible region. C If b = 0, check the sign of a i. If a < 0 move the ruler to the “leftmost” feasible point. This point is then the solution. ii. If a > 0 move the ruler to the “rightmost” feasible point. This point is then the solution.

Worked Example 188: Prizes! Question: As part of their opening specials, a furniture store has promised to give away at least 40 prizes with a total value of at least R2 000. The prizes are kettles and toasters. 1. If the company decides that there will be at least 10 of each prize, write down two more inequalities from these constraints. 2. If the cost of manufacturing a kettle is R60 and a toaster is R50, write down an objective function C which can be used to determine the cost to the company of both kettles and toasters. 542

CHAPTER 41. LINEAR PROGRAMMING - GRADE 12 3. Sketch the graph of the feasibility region that can be used to determine all the possible combinations of kettles and toasters that honour the promises of the company. 4. How many of each prize will represent the cheapest option for the company? 5. How much will this combination of kettles and toasters cost? Answer Step 1 : Identify the decision variables Let the number of kettles be xk and the number of toasters be yt and write down two constraints apart from xk ≥ 0 and yt ≥ 0 that must be adhered to. Step 2 : Write constraint equations Since there will be at least 10 of each prize we can write: xk ≥ 10 and yt ≥ 10 Also the store has promised to give away at least 40 prizes in total. Therefore: xk + yt ≥ 40 Step 3 : Write the objective function The cost of manufacturing a kettle is R60 and a toaster is R50. Therefore the cost the total cost C is: C = 60xk + 50yt Step 4 : Sketch the graph of the feasible region yt 100 90 80 70 60 50 40 30 20 10 10 20 A xk 30 40 50 60 70 80 90 100 B

41.3

Step 5 : Determine vertices of feasible region From the graph, the coordinates of vertex A is (3,1) and the coordinates of vertex B are (1,3). Step 6 : Draw in the search line The seach line is the gradient of the objective function. That is, if the equation C = 60x + 50y is now written in the standard form y = ..., then the gradient is: 6 m=− , 5 which is shown with the broken line on the graph. 543

Step 8 : Write the ﬁnal answer The cheapest combination of prizes is 10 kettles and 30 toasters, costing the company R2 100.

Worked Example 189: Search Line Method Question: As a production planner at a factory manufacturing lawn cutters your job will be to advise the management on how many of each model should be produced per week in order to maximise the proﬁt on the local production. The factory is producing two types of lawn cutters: Quadrant and Pentagon. Two of the production processes that the lawn cutters must go through are: bodywork and engine work. • The factory cannot operate for less than 360 hours on engine work for the lawn cutters. • The factory has a maximum capacity of 480 hours for bodywork for the lawn cutters. 544

CHAPTER 41. LINEAR PROGRAMMING - GRADE 12 • Half an hour of engine work and half an hour of bodywork is required to produce one Quadrant. • One third of an hour of engine work andone ﬁfth of an hour of bodywork is required to produce one Pentagon. • The ratio of Pentagon lawn cutters to Quadrant lawn cutters produced per week must be at least 3:2. Let the number of Quadrant lawn cutters manufactured in a week be x. Let the number of Pentagon lawn cutters manufactured in a week be y. Two of the constraints are: x ≥ 200 3x + 2y ≥ 2 160 1. Write down the remaining constraints in terms of x and y to represent the abovementioned information. 2. Use graph paper to represent the constraints graphically. 3. Clearly indicate the feasible region by shading it. 4. If the proﬁt on one Quadrant lawn cutter is R1 200 and the proﬁt on one Pentagon lawn cutter is R400, write down an equation that will represent the proﬁt on the lawn cutters. 5. Using a search line and your graph, determine the number of Quadrant and Pentagon lawn cutters that will yield a maximum proﬁt. 6. Determine the maximum proﬁt per week. Answer Step 1 : Remaining constraints: 1 1 x + ≤ 480 2 5 3 y ≥ x 2 Step 2 : Graphical representation y 2400 • A minimum of 200 Quadrant lawn cutters must be produced per week.

1. Polkadots is a small company that makes two types of cards, type X and type Y. With the available labour and material, the company can make not more than 150 cards of type X and not more than 120 cards of type Y per week. Altogether they cannot make more than 200 cards per week. There is an order for at least 40 type X cards and 10 type Y cards per week. Polkadots makes a proﬁt of R5 for each type X card sold and R10 for each type Y card. Let the number of type X cards be x and the number of type Y cards be y, manufactured per week. A One of the constraint inequalities which represents the restrictions above is x ≤ 150. Write the other constraint inequalities. B Represent the constraints graphically and shade the feasible region. C Write the equation that represents the proﬁt P (the objective function), in terms of x and y. D On your graph, draw a straight line which will help you to determine how many of each type must be made weekly to produce the maximum P E Calculate the maximum weekly proﬁt. 2. A brickworks produces “face bricks” and “clinkers”. Both types of bricks are produced and sold in batches of a thousand. Face bricks are sold at R150 per thousand, and clinkers at R100 per thousand, where an income of at least R9,000 per month is required to cover costs. The brickworks is able to produce at most 40,000 face bricks and 90,000 clinkers per month, and has transport facilities to deliver at most 100,000 bricks per month. The number of clinkers produced must be at least the same number of face bricks produced. Let the number of face bricks in thousands be x, and the number of clinkers in thousands be y. A List all the constraints. B Graph the feasible region. C If the sale of face bricks yields a proﬁt of R25 per thousand and clinkers R45 per thousand, use your graph to determine the maximum proﬁt. D If the proﬁt margins on face bricks and clinkers are interchanged, use your graph to determine the maximum proﬁt. 3. A small cell phone company makes two types of cell phones: Easyhear and Longtalk. Production ﬁgures are checked weekly. At most, 42 Easyhear and 60 Longtalk phones can be manufactured each week. At least 30 cell phones must be produced each week to cover costs. In order not to ﬂood the market, the number of Easyhear phones cannot be more than twice the number of Longtalk phones. It takes 2 hour to assemble an Easyhear 3 phone and 1 hour to put together a Longtalk phone. The trade unions only allow for a 2 50-hour week. Let x be the number of Easyhear phones and y be the number of Longtalk phones manufactured each week. A Two of the constraints are: 0 ≤ x ≤ 42 and 0 ≤ y ≤ 60

C If the proﬁt on an Easyhear phone is R225 and the proﬁt on a Longtalk is R75, determine the maximum proﬁt per week. 4. Hair for Africa is a ﬁrm that specialises in making two kinds of up-market shampoo, Glowhair and Longcurls. They must produce at least two cases of Glowhair and one case of Longcurls per day to stay in the market. Due to a limited supply of chemicals, they cannot produce more than 8 cases of Glowhair and 6 cases of Longcurls per day. It takes half-an-hour to produce one case of Glowhair and one hour to produce a case of Longcurls, and due to restrictions by the unions, the plant may operate for at most 7 hours per day. The workforce at Hair for Africa, which is still in training, can only produce a maximum of 10 cases of shampoo per day. Let x be the number of cases of Glowhair and y the number of cases of Longcurls produced per day. A Write down the inequalities that represent all the constraints. B Sketch the feasible region. C If the proﬁt on a case of Glowhair is R400 and the proﬁt on a case of Longcurls is R300, determine the maximum proﬁt that Hair for Africa can make per day. 5. A transport contracter has 6 5-ton trucks and 8 3-ton trucks. He must deliver at least 120 tons of sand per day to a construction site, but he may not deliver more than 180 tons per day. The 5-ton trucks can each make three trips per day at a cost of R30 per trip, and the 3-ton trucks can each make four trips per day at a cost of R120 per trip. How must the contracter utilise his trucks so that he has minimum expense ?

547

41.4

CHAPTER 41. LINEAR PROGRAMMING - GRADE 12

548

Chapter 42

Geometry - Grade 12
42.1 Introduction

Activity :: Discussion : Discuss these Research Topics Research one of the following geometrical ideas and describe it to your group: 1. taxicab geometry, 2. sperical geometry, 3. fractals, 4. the Koch snowﬂake.

42.2
42.2.1

Circle Geometry
Terminology

The following is a recap of terms that are regularly used when referring to circles. arc An arc is a part of the circumference of a circle. chord A chord is deﬁned as a straight line joining the ends of an arc. radius The radius, r, is the distance from the centre of the circle to any point on the circumference. diameter The diameter, , is a special chord that passes through the centre of the circle. The diameter is the straight line from a point on the circumference to another point on the circumference, that passes through the centre of the circle. segment A segment is the part of the circle that is cut oﬀ by a chord. A chord divides a circle into two segments. tangent A tangent is a line that makes contact with a circle at one point on the circumference. (AB is a tangent to the circle at point P . 549

42.2

CHAPTER 42. GEOMETRY - GRADE 12

segment chord

rad

i us

O diameter

A

a rc
P tangent B C O P 550 B

Figure 42.1: Parts of a Circle

42.2.2

Axioms

An axiom is an established or accepted principle. For this section, the following are accepted as axioms. 1. The Theorem of Pythagoras, which states that the square on the hypotenuse of a rightangled triangle is equal to the sum of the squares on the other two sides. In △ABC, this means that AB 2 + BC 2 = AC 2 A

B

Figure 42.2: A right-angled triangle 2. A tangent is perpendicular to the radius, drawn at the point of contact with the circle.

42.2.3

Theorems of the Geometry of Circles

A theorem is a general proposition that is not self-evident but is proved by reasoning (these proofs need not be learned for examination purposes). Theorem 6. The line drawn from the centre of a circle, perpendicular to a chord, bisects the chord. Proof:

A

CHAPTER 42. GEOMETRY - GRADE 12

42.2

Consider a circle, with centre O. Draw a chord AB and draw a perpendicular line from the centre of the circle to intersect the chord at point P . The aim is to prove that AP = BP 1. △OAP and △OBP are right-angled triangles. 2. OA = OB as both of these are radii and OP is common to both triangles. Apply the Theorem of Pythagoras to each triangle, to get: OA2 OB 2 However, OA = OB. So, OP 2 + AP 2 ∴ AP 2 and AP This means that OP bisects AB. Theorem 7. The line drawn from the centre of a circle, that bisects a chord, is perpendicular to the chord. Proof: = = = OP 2 + BP 2 BP 2 BP = OP 2 + AP 2 = OP 2 + BP 2

Theorem 8. The perpendicular bisector of a chord passes through the centre of the circle.

Proof:

Q

A

P

B

Consider a circle. Draw a chord AB. Draw a line P Q perpendicular to AB such that P Q bisects AB at point P . Draw lines AQ and BQ. The aim is to prove that Q is the centre of the circle, by showing that AQ = BQ. In △OAP and △OBP ,

1. AP = P B (given)

2. ∠QP A = ∠QP B (QP ⊥ AB)

3. QP is common to both triangles.

∴ △QAP ≡ △QBP (SAS). From this, QA = QB. Since the centre of a circle is the only point inside a circle that has points on the circumference at an equal distance from it, Q must be the centre of the circle.

Exercise: Circles I

1. Find the value of x: 552

CHAPTER 42. GEOMETRY - GRADE 12

42.2

a)

b)

O x 5 Q P PR=8 x R P

O 4 Q R

PR=6

c)
10 O x R P Q PR=8

d)

Q 2 6 P x O S 6

R

e)
S 5 P 8 10 O x T U R Q

f)
24 T P 5 x 25 S

Q

R O

R

Theorem 9. The angle subtended by an arc at the centre of a circle is double the size of the angle subtended by the same arc at the circumference of the circle. Proof: P

Theorem 10. The angles subtended by a chord at the circumference of a circle on the same side of the chord are equal. Proof: Q P

O A

B Consider a circle, with centre O. Draw a chord AB. Select any points P and Q on the circumference of the circle, such that both P and Q are on the same side of the chord. Draw lines P A, P B, QA and QB. 554

Theorem 11. (Converse of Theorem 10) If a line segment subtends equal angles at two other points on the same side of the line, then these four points lie on a circle. Proof: Q P R

A

B

Consider a line segment AB, that subtends equal angles at points P and Q on the same side of AB. The aim is to prove that points A, B, P and Q lie on the circumference of a circle. By contradiction. Assume that point P does not lie on a circle drawn through points A, B and Q. Let the circle cut AP (or AP extended) at point R.

∴ the assumption that the circle does not pass through P , must be false, and A, B, P and Q lie on the circumference of a circle.

Exercise: Circles III 1. Find the values of the unknown letters. 555

42.2 1. A
a

CHAPTER 42. GEOMETRY - GRADE 12 2. E

B
21◦

F
15◦

D G C 3. J 4. N
b

I

H

O K
c 17◦

M Q
24◦ d

L

P

5.

T

S R
45◦ 35◦

6.
35◦

W

X O

12◦ e

f

Y V

Z

U

Cyclic Quadrilaterals

Cyclic quadrilaterals are quadrilaterals with all four vertices lying on the circumference of a circle. The vertices of a cyclic quadrilateral are said to be concyclic. Theorem 12. The opposite angles of a cyclic quadrilateral are supplementary.

Theorem 13. (Converse of Theorem 12) If the opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic. Proof: Q R P

A

B ˆ ˆ ˆ ˆ Consider a quadrilateral ABP Q, such that ABP + AQP = 180◦ and QAB + QP B = 180◦ . The aim is to prove that points A, B, P and Q lie on the circumference of a circle. By contradiction. Assume that point P does not lie on a circle drawn through points A, B and Q. Let the circle cut AP (or AP extended) at point R. Draw BR.

Theorem 14. Two tangents drawn to a circle from the same point outside the circle are equal in length. Proof:

A

O

P B Consider a circle, with centre O. Choose a point P outside the circle. Draw two tangents to the circle from point P , that meet the circle at A and B. Draw lines OA, OB and OP . The aim is to prove that AP = BP . In △OAP and △OBP , 558

Theorem 16. (Converse of 15) If the angle formed between a line, that is drawn through the end point of a chord, and the chord, is equal to the angle subtended by the chord in the alternate segment, then the line is a tangent to the circle. Proof:

Q A O

Y S X B 561 R

42.2

CHAPTER 42. GEOMETRY - GRADE 12

Consider a circle, with centre O and chord AB. Let line SR pass through point B. Chord AB ˆ ˆ subtends an angle at point Q such that ABS = AQB. The aim is to prove that SBR is a tangent to the circle. By contradiction. Assume that SBR is not a tangent to the circle and draw XBY such that XBY is a tangent to the circle.

However,

ˆ ABX ˆ ABS

= = = = =

ˆ AQB ˆ AQB

(tan-chord theorem) (given) (42.2)

ˆ ∴ ABX ˆ But since, ABX ˆ (42.2) can only be true if, XBS

ˆ ABS ˆ ˆ ABS + XBS 0

ˆ If XBS is zero, then both XBY and SBR coincide and SBR is a tangent to the circle.

Exercise: Applying Theorem 9

1. Show that Theorem 9 also applies to the following two cases:

A

P

O

O R P

R

B A B

562

CHAPTER 42. GEOMETRY - GRADE 12

42.2

Worked Example 190: Circle Geometry I BD is a tangent to the circle with centre O. BO ⊥ AD. Prove that: 1. CF OE is a cyclic quadrilateral
O A E D

Question:
F C

2. F B = BC 3. △COE///△CBF 4. CD2 = ED.AD 5.
OE BC

=

CD CO

B

Answer 1. Step 1 : To show a quadrilateral is cyclic, we need a pair of opposite angles to be supplementary, so lets look for that. ˆ F OE ˆ F CE

2. Step 1 : Since these two sides are part of a triangle, we are proving that triangle to be isosceles. The easiest way is to show the angles opposite to those sides to be equal. ˆ Let OEC = x. ∴ ∴ ∴ ˆ F CB = x (∠ between tangent BD and chord CE) ˆ BF C = x (exterior ∠ to cyclic quadrilateral CF OE) BF = BC (sides opposite equal ∠’s in isosceles △BF C)

3. Step 1 : To show these two triangles similar, we will need 3 equal angles. We already have 3 of the 6 needed angles from the previous question. We need only ﬁnd the missing 3 angles. ˆ CBF OC

Step 3 : The third equal angle is an angle both triangles have in common. ˆ ˆ Lastly, ADC = EDC since they are the same ∠. Step 4 : Now we know that the triangles are similar and can use the proportionality relation accordingly.

∴ △ADC///△CDE (3 ∠’s equal) ED CD ∴ = CD AD ∴ CD2 = ED.AD 5. Step 1 : This looks like another proportionality relation with a little twist, since not all sides are contained in 2 triangles. There is a quick observation we can make about the odd side out, OE.

OE

=

CD (△OEC is isosceles)

Step 2 : With this observation we can limit ourselves to proving triangles BOC and ODC similar. Start in one of the triangles. In △BCO ˆ OCB ˆ CBO = = 90◦ (radius OC on tangent BD) 180◦ − 2x (sum of ∠’s in △BF C)

Step 3 : Then we move on to the other one. In △OCD ˆ OCD ˆ COD = = 90◦ (radius OC on tangent BD) 180◦ − 2x (sum of ∠’s in △OCE)

Step 4 : Again we have a common element. Lastly, OC is a common side to both △’s. Step 5 : Then, once we’ve shown similarity, we use the proportionality relation , as well as our ﬁrst observation, appropriately. 564

Step 2 : We have already proved 1 pair of angles equal in the previous question.

∠BCD

= 565

∠F AE (above)

42.3

CHAPTER 42. GEOMETRY - GRADE 12 Step 3 : Proving the last set of angles equal is simply a matter of adding up the angles in the triangles. Then we have proved similarity.

∠AF E ∠CBD

= = ∴

△AF E///△CBD (3 ∠’s equal)

180◦ − x − y (∠’s in △AF E) 180◦ − x − y (∠’s in △CBD)

3. Step 1 : This equation looks like it has to do with proportionality relation of similar triangles. We already showed triangles AF E and CBD similar in the previous question. So lets start there. DC BD FA FE DC.F E = FA BD

= ∴

Step 2 : Now we need to look for a hint about side F A. Looking at triangle CAH we see that there is a line F G intersecting it parallel to base CH. This gives us another proportionality relation. AG GH FA (F G CH splits up lines AH and AC proportionally) FC F C.AG FA = GH

We know that every point on the circumference of a circle is the same distance away from the centre of the circle. Consider a point (x1 ,y1 ) on the circumference of a circle of radius r with centre at (x0 ,y0 ).

P (x1 ,y1 ) (x0 ,y0 ) O Q

Figure 42.3: Circle h with centre (x0 ,y0 ) has a tangent, g passing through point P at (x1 ,y1 ). Line f passes through the centre and point P . 566

Worked Example 192: Equation of a Circle I Question: Find the equation of a circle (centre O) with a diameter between two points, P at (−5,5) and Q at (5, − 5).

Answer

Step 1 : Draw a picture Draw a picture of the situation to help you ﬁgure out what needs to be done.

P

5 O

−5

5 −5 Q

Step 2 : Find the centre of the circle We know that the centre of a circle lies on the midpoint of a diameter. Therefore the co-ordinates of the centre of the circle is found by ﬁnding the midpoint of the line between P and Q. Let the co-ordinates of the centre of the circle be (x0 ,y0 ), let the co-ordinates of P be (x1 ,y1 ) and let the co-ordinates of Q be (x2 ,y2 ). Then, 567

We are given that a tangent to a circle is drawn through a point P with co-ordinates (x1 ,y1 ). In this section, we ﬁnd out how to determine the equation of that tangent.

g

h f (x0 ,y0 )

P (x1 ,y1 )

Figure 42.4: Circle h with centre (x0 ,y0 ) has a tangent, g passing through point P at (x1 ,y1 ). Line f passes through the centre and point P .

We start by making a list of what we know: 1. We know that the equation of the circle with centre (x0 ,y0 ) is (x − x0 )2 + (y − y0 )2 = r2 . 2. We know that a tangent is perpendicular to the radius, drawn at the point of contact with the circle. As we have seen in earlier grades, there are two steps to determining the equation of a straight line: Step 1: Calculate the gradient of the line, m. Step 2: Calculate the y-intercept of the line, c. The same method is used to determine the equation of the tangent. First we need to ﬁnd the gradient of the tangent. We do this by ﬁnding the gradient of the line that passes through the centre of the circle and point P (line f in Figure 42.4), because this line is a radius line and the tangent is perpendicular to it. mf = y1 − y0 x1 − x0 (42.4)

For example, ﬁnd the equation of the tangent to the circle at point (1,1). The centre of the circle is at (0,0). The equation of the circle is x2 + y 2 = 2. Use y − y1 = − with (x0 ,y0 ) = (0,0) and (x1 ,y1 ) = (1,1). x1 − x0 (x − x1 ) y1 − y0

Exercise: Rotations Any line OP is drawn (not necessarily in the ﬁrst quadrant), making an angle of θ degrees with the x-axis. Using the co-ordinates of P and the angle α, calculate the co-ordinates of P ′ , if the line OP is rotated about the origin through α degrees. 1. 2. 3. 4. 5. 6. P (2, 6) (4, 2) (5, -1) (-3, 2) (-4, -1) (2, 5) α 60◦ 30◦ 45◦ 120◦ 225◦ -150◦ O

P

θ

42.4.2

Characteristics of Transformations

Rigid transformations like translations, reﬂections, rotations and glide reﬂections preserve shape and size, and that enlargement preserves shape but not size.

42.4.3

Characteristics of Transformations

Rigid transformations like translations, reﬂections, rotations and glide reﬂections preserve shape and size, and that enlargement preserves shape but not size.

Activity :: : Geometric Transformations
15

10

Draw a large 15×15 grid and plot △ABC with A(2; 6), B(5; 6) and C(5; 1). Fill in the lines y = x and y = −x. Complete the table below , by drawing the images of △ABC under the given transformations. The ﬁrst one has been done for you.
−15 −10 −5

We have, for any angles α and β, that sin(α + β) = sin α cos β + sin β cos α How do we derive this identity? It is tricky, so follow closely. Suppose we have the unit circle shown below. The two points L(a,b) and K(x,y) are on the circle.
y

K(x; y) L(a; b) 1 (α − β) α β
a

1

b

O

M (x; y)

x

We can get the coordinates of L and K in terms of the angles α and β. For the triangle LOK, we have that b 1 a cos β = 1 sin β = =⇒ =⇒ 577 b = sin β a = cos β

The most important thing to remember when asked to prove identities is:

Important: Trigonometric Identities

Never assume that the left hand side is equal to the right hand side. You need to show that both sides are equal. A suggestion for proving identities: It is usually much easier simplifying the more complex side of an identity to get the simpler side than the other way round.

sin θ + sin 2θ = tan θ 1 + cos θ + cos 2θ For which values is the identity not valid? Answer Step 1 : Identify a strategy The right-hand side (RHS) of the identity cannot be simpliﬁed. Thus we should try simplify the left-hand side (LHS). We can also notice that the trig function on the RHS does not have a 2θ dependance. Thus we will need to use the doubleangle formulas to simplify the sin 2θ and cos 2θ on the LHS. We know that tan θ is undeﬁned for some angles θ. Thus the identity is also undeﬁned for these θ, and hence is not valid for these angles. Also, for some θ, we might have division by zero in the LHS, which is not allowed. Thus the identity won’t hold for these angles also. Step 2 : Execute the strategy sin θ + 2 sin θ cos θ 1 + cos θ + (2 cos2 θ − 1) sin θ(1 + 2 cos θ) cos θ(1 + 2 cos θ) sin θ cos θ tan θ RHS

Worked Example 199: Height of tower Question: D is the top of a tower of height h. Its base is at C. The triangle ABC lies on ˆ ˆ the ground (a horizontal plane). If we have that BC = b, DBA = α, DBC = x ˆ and DCB = θ, show that b sin α sin x h= sin(x + θ) 584

CHAPTER 43. TRIGONOMETRY - GRADE 12
D

43.2

h

C θ

A

b x B

α

Answer Step 1 : Identify a strategy We have that the triangle ABD is right-angled. Thus we can relate the height h with the angle α and either the length BA or BD (using sines or cosines). But we have two angles and a length for △BCD, and thus can work out all the remaining lengths and angles of this triangle. We can thus work out BD. Step 2 : Execute the strategy We have that h BD =⇒ h = = sin α BD sin α

Exercise: 1. The line BC represents a tall tower, with C at its foot. Its angle of elevation from D is θ. We are also given that BA = AD = x. 585

43.3 C

CHAPTER 43. TRIGONOMETRY - GRADE 12

B

θ α x A x

D

A Find the height of the tower BC in terms of x, tan θ and cos 2α. B Find BC if we are given that k = 140m, α = 21◦ and θ = 9◦ .

43.3
43.3.1

Other Geometries
Taxicab Geometry

Taxicab geometry, considered by Hermann Minkowski in the 19th century, is a form of geometry in which the usual metric of Euclidean geometry is replaced by a new metric in which the distance between two points is the sum of the (absolute) diﬀerences of their coordinates.

43.3.2

Manhattan distance

The metric in taxi-cab geometry, is known as the Manhattan distance, between two points in an Euclidean space with ﬁxed Cartesian coordinate system as the sum of the lengths of the projections of the line segment between the points onto the coordinate axes. For example, in the plane, the Manhattan distance between the point P1 with coordinates (x1 , y1 ) and the point P2 at (x2 , y2 ) is |x1 − x2 | + |y1 − y2 | (43.1)

Figure 43.1: Manhattan Distance (dotted and solid) compared to Euclidean Distance (dashed). √ In each case the Manhattan distance is 12 units, while the Euclidean distance is 36 586

CHAPTER 43. TRIGONOMETRY - GRADE 12

43.3

The Manhattan distance depends on the choice on the rotation of the coordinate system, but does not depend on the translation of the coordinate system or its reﬂection with respect to a coordinate axis. Manhattan distance is also known as city block distance or taxi-cab distance. It is given these names because it is the shortest distance a car would drive in a city laid out in square blocks. Taxicab geometry satisﬁes all of Euclid’s axioms except for the side-angle-side axiom, as one can generate two triangles with two sides and the angle between them the same and have them not be congruent. In particular, the parallel postulate holds. A circle in taxicab geometry consists of those points that are a ﬁxed Manhattan distance from the center. These circles are squares whose sides make a 45◦ angle with the coordinate axes.

43.3.3

Spherical Geometry

Spherical geometry is the geometry of the two-dimensional surface of a sphere. It is an example of a non-Euclidean geometry. In plane geometry the basic concepts are points and line. On the sphere, points are deﬁned in the usual sense. The equivalents of lines are not deﬁned in the usual sense of ”straight line” but in the sense of ”the shortest paths between points” which is called a geodesic. On the sphere the geodesics are the great circles, so the other geometric concepts are deﬁned like in plane geometry but with lines replaced by great circles. Thus, in spherical geometry angles are deﬁned between great circles, resulting in a spherical trigonometry that diﬀers from ordinary trigonometry in many respects (for example, the sum of the interior angles of a triangle exceeds 180◦). Spherical geometry is the simplest model of elliptic geometry, in which a line has no parallels through a given point. Contrast this with hyperbolic geometry, in which a line has two parallels, and an inﬁnite number of ultra-parallels, through a given point. Spherical geometry has important practical uses in celestial navigation and astronomy.

Extension: Distance on a Sphere The great-circle distance is the shortest distance between any two points on the surface of a sphere measured along a path on the surface of the sphere (as opposed to going through the sphere’s interior). Because spherical geometry is rather diﬀerent from ordinary Euclidean geometry, the equations for distance take on a diﬀerent form. The distance between two points in Euclidean space is the length of a straight line from one point to the other. On the sphere, however, there are no straight lines. In non-Euclidean geometry, straight lines are replaced with geodesics. Geodesics on the sphere are the great circles (circles on the sphere whose centers are coincident with the center of the sphere). Between any two points on a sphere which are not directly opposite each other, there is a unique great circle. The two points separate the great circle into two arcs. The length of the shorter arc is the great-circle distance between the points. Between two points which are directly opposite each other (called antipodal points) there inﬁnitely many great circles, but all have the same length, equal to half the circumference of the circle, or πr, where r is the radius of the sphere. Because the Earth is approximately spherical (see spherical Earth), the equations for great-circle distance are important for ﬁnding the shortest distance between points on the surface of the Earth, and so have important applications in navigation. Let φ1 ,λ1 ; φ2 ,λ2 , be the latitude and longitude of two points, respectively. Let ∆λ be the longitude diﬀerence. Then, if r is the great-circle radius of the sphere, the great-circle distance is r∆σ, where ∆σ is the angular diﬀerence/distance and can be determined from the spherical law of cosines as: ∆σ = arccos {sin φ1 sin φ2 + cos φ1 cos φ2 cos ∆λ}

587

43.3

CHAPTER 43. TRIGONOMETRY - GRADE 12 Extension: Spherical Distance on the Earth The shape of the Earth more closely resembles a ﬂattened spheroid with extreme values for the radius of curvature, or arcradius, of 6335.437 km at the equator (vertically) and 6399.592 km at the poles, and having an average great-circle radius of 6372.795 km. Using a sphere with a radius of 6372.795 km thus results in an error of up to about 0.5%.

43.3.4

Fractal Geometry

The word ”fractal” has two related meanings. In colloquial usage, it denotes a shape that is recursively constructed or self-similar, that is, a shape that appears similar at all scales of magniﬁcation and is therefore often referred to as ”inﬁnitely complex.” In mathematics a fractal is a geometric object that satisﬁes a speciﬁc technical condition, namely having a Hausdorﬀ dimension greater than its topological dimension. The term fractal was coined in 1975 by Benot Mandelbrot, from the Latin fractus, meaning ”broken” or ”fractured.” Three common techniques for generating fractals are:

• Escape-time fractals - Fractals deﬁned by a recurrence relation at each point in a space (such as the complex plane). Examples of this type are the Mandelbrot set, the Burning Ship fractal and the Lyapunov fractal.

Fractals in nature Approximate fractals are easily found in nature. These objects display self-similar structure over an extended, but ﬁnite, scale range. Examples include clouds, snow ﬂakes, mountains, river networks, and systems of blood vessels. Trees and ferns are fractal in nature and can be modeled on a computer using a recursive algorithm. This recursive nature is clear in these examples - a branch from a tree or a frond from a fern is a miniature replica of the whole: not identical, but similar in nature. The surface of a mountain can be modeled on a computer using a fractal: Start with a triangle in 3D space and connect the central points of each side by line segments, resulting in 4 triangles. The central points are then randomly moved up or down, within a deﬁned range. The procedure is repeated, decreasing at each iteration the range by half. The recursive nature of the algorithm guarantees that the whole is statistically similar to each detail. 588

In this chapter, you will use the mean, median, mode and standard deviation of a set of data to identify whether the data is normally distributed or whether it is skewed. You will learn more about populations and selecting diﬀerent kinds of samples in order to avoid bias. You will work with lines of best ﬁt, and learn how to ﬁnd a regression equation and a correlation coeﬃcient. You will analyse these measures in order to draw conclusions and make predictions.

1. Calculate the mean, median, mode and standard deviation of the data. 2. What percentage of the data is within one standard deviation of the mean? 3. Draw a histogram of the data using intervals 60 ≤ x < 64, 64 ≤ x < 68, etc. 4. Join the midpoints of the bars to form a frequency polygon.

If large numbers of data are collected from a population, the graph will often have a bell shape. If the data was, say, examination results, a few learners usually get very high marks, a few very low marks and most get a mark in the middle range. We say a distribution is normal if • the mean, median and mode are equal. • it is symmetric around the mean. • ±68% of the sample lies within one standard deviation of the mean, 95% within two standard deviations and 99% within three standard deviations of the mean. 591

44.2

CHAPTER 44. STATISTICS - GRADE 12

68% 95% 99%
¯ x − 3σ x − 2σ x − σ ¯ ¯ x ¯ x + σ x + 2σ x + 3σ ¯ ¯ ¯

What happens if the test was very easy or very diﬃcult? Then the distribution may not be symmetrical. If extremely high or extremely low scores are added to a distribution, then the mean tends to shift towards these scores and the curve becomes skewed.

If the test was very diﬃcult, the mean score is shifted to the left. In this case, we say the distribution is positively skewed, or skewed right. Skewed right If it was very easy, then many learners would get high scores, and the mean of the distribution would be shifted to the right. We say the distribution is negatively skewed, or skewed left. Skewed left

Draw the histogram of the results. Join the midpoints of each bar and draw a frequency polygon. What mark must one obtain in order to be in the top 2% of the class? Approximately 84% of the pupils passed the test. What was the pass mark? Is the distribution normal or skewed? 592

CHAPTER 44. STATISTICS - GRADE 12 3. In a road safety study, the speed of 175 cars was monitored along a speciﬁc stretch of highway in order to ﬁnd out whether there existed any link between high speed and the large number of accidents along the route. A frequency table of the results is drawn up below. Speed (km.h−1 ) 50 60 70 80 90 100 110 120 Number of cars (Frequency) 19 28 23 56 20 16 8 5

44.3

The mean speed was determined to be around 82 km.h−1 while the median speed was worked out to be around 84,5 km.h−1 . A Draw a frequency polygon to visualise the data in the table above. B Is this distribution symmetrical or skewed left or right? Give a reason fro your answer.

44.3

Extracting a Sample Population

Suppose you are trying to ﬁnd out what percentage of South Africa’s population owns a car. One way of doing this might be to send questionnaires to peoples homes, asking them whether they own a car. However, you quickly run into a problem: you cannot hope to send every person in the country a questionnaire, it would be far to expensive. Also, not everyone would reply. The best you can do is send it to a few people, see what percentage of these own a car, and then use this to estimate what percentage of the entire country own cars. This smaller group of people is called the sample population. The sample population must be carefully chosen, in order to avoid biased results. How do we do this? First, it must be representative. If all of our sample population comes from a very rich area, then almost all will have cars. But we obviously cannot conclude from this that almost everyone in the country has a car! We need to send the questionnaire to rich as well as poor people. Secondly, the size of the sample population must be large enough. It is no good having a sample population consisting of only two people, for example. Both may very well not have cars. But we obviously cannot conclude that no one in the country has a car! The larger the sample population size, the more likely it is that the statistics of our sample population corresponds to the statistics of the entire population. So how does one ensure that ones sample is representative? There are a variety of methods available, which we will look at now.

Random Sampling. Every person in the country has an equal chance of being selected. It is unbiased and also independant, which means that the selection of one person has no eﬀect on the selection on another. One way of doing this would be to give each person in the country a number, and then ask a computer to give us a list of random numbers. We could then send the questionnaire to the people corresponding to the random numbers. Systematic Sampling. Again give every person in the country a number, and then, for example, select every hundredth person on the list. So person with number 1 would be selected, person with number 100 would be selected, person with number 200 would be selected, etc. 593

44.4

CHAPTER 44. STATISTICS - GRADE 12 Stratiﬁed Sampling. We consider diﬀerent subgroups of the population, and take random samples from these. For example, we can divide the population into male and female, diﬀerent ages, or into diﬀerent income ranges. Cluster Sampling. Here the sample is concentrated in one area. For example, we consider all the people living in one urban area.

Exercise: Sampling 1. Discuss the advantages, disadvantages and possible bias when using A systematic sampling B random sampling C cluster sampling 2. Suggest a suitable sampling method that could be used to obtain information on: A passengers views on availability of a local taxi service. B views of learners on school meals. C defects in an item made in a factory. D medical costs of employees in a large company. 3. 5% of a certain magazines’ subscribers is randomly selected. The random number 16 out of 50, is selected. Then subscribers with numbers 16, 66, 116, 166, . . . are chosen as a sample. What kind of sampling is this?

44.4

Function Fitting and Regression Analysis

In Grade 11 we recorded two sets of data (bivariate data) on a scatter plot and then we drew a line of best ﬁt as close to as many of the data items as possible. Regression analysis is a method of ﬁnding out exactly which function best ﬁts a given set of data. We can ﬁnd out the equation of the regression line by drawing and estimating, or by using an algebraic method called “the least squared method”, or we can use a calculator. The linear regression equation is written y = a + bx (we say y-hat) or y = A + Bx. Of course these are both variations of a more familiar ˆ equation y = mx + c. Suppose you are doing an experiment with washing dishes. You count how many dishes you begin with, and then ﬁnd out how long it takes to ﬁnish washing them. So you plot the data on a graph of time taken versus number of dishes. This is plotted below.

t 200 180

Time taken (seconds)

160 140 120 100 80 60 40 20 0 0 1 2 3 4 5 6 Number of dishes d

594

CHAPTER 44. STATISTICS - GRADE 12

44.4

If t is the time taken, and d the number of dishes, then it looks as though t is proportional to d, ie. t = m · d, where m is the constant of proportionality. There are two questions that interest us now. 1. How do we ﬁnd m? One way you have already learnt, is to draw a line of best-ﬁt through the data points, and then measure the gradient of the line. But this is not terribly precise. Is there a better way of doing it? 2. How well does our line of best ﬁt really ﬁt our data? If the points on our plot don’t all lie close to the line of best ﬁt, but are scattered everywhere, then the ﬁt is not ’good’, and our assumption that t = m · d might be incorrect. Can we ﬁnd a quantitative measure of how well our line really ﬁts the data? In this chapter, we answer both of these questions, using the techniques of regression analysis.

Worked Example 200: Fitting by hand Question: Use the data given to draw a scatter plot and line of best ﬁt. Now write down the equation of the line that best seems to ﬁt the data. x y 1,0 2,5 2,4 2,8 3,1 3,0 4,9 4,8 5,6 5,1 6,2 5,3

Step 2 : Calculating the equation of the line The equation of the line is y = mx + c From the graph we have drawn, we estimate the y-intercept to be 1,5. We estimate that y = 3,5 when x = 3. So we have that points (3; 3,5) and (0; 1,6) lie on the line. The gradient of the line, m, is given by m = = = y2 − y1 x2 − x1 3,5 − 1,5 3−0 2 3

So we ﬁnally have that the equation of the line of best ﬁt is y= 2 x + 1,5 3 595

44.4

CHAPTER 44. STATISTICS - GRADE 12

44.4.1

The Method of Least Squares

We now come to a more accurate method of ﬁnding the line of best-ﬁt. The method is very simple. Suppose we guess a line of best-ﬁt. Then at at every data point, we ﬁnd the distance between the data point and the line. If the line ﬁtted the data perfectly, this distance should be zero for all the data points. The worse the ﬁt, the larger the diﬀerences. We then square each of these distances, and add them all together. y

y

The best-ﬁt line is then the line that minimises the sum of the squared distances. Suppose we have a data set of n points {(x1 ; y1 ), (x2 ; y2 ), . . . , (xn ,yn )}. We also have a line f (x) = mx + c that we are trying to ﬁt to the data. The distance between the ﬁrst data point and the line, for example, is distance = y1 − f (x) = y1 − (mx + c) We now square each of these distances and add them together. Lets call this sum S(m,c). Then we have that S(m,c) = =
i=1

Worked Example 201: Method of Least Squares Question: In the table below, we have the records of the maintenance costs in Rands, compared with the age of the appliance in months. We have data for 5 appliances.

Now press [2] for linear regression. Your screen should look something like this: x 1 2 3 Step 2 : Entering the data Press [52] and then [=] to enter the ﬁrst mark under x. Then enter the other values, in the same way, for the x-variable (the Chemistry marks) in the order in which they are given in the data set. Then move the cursor across and up and enter 48 under y opposite 52 in the x-column. Continue to enter the other y-values (the Accounting marks) in order so that they pair oﬀ correctly with the corresponding x-values. 1 2 3 x 52 55 y y

Step 3 : Getting regression results from the calculator a) Press [1] and [=] to get the value of the y-intercept, a = −5,065.. = −5,07(to 2 d.p.) Finally, to get the slope, use the following key sequence: [SHIFT][1][7][2][=]. The calculator gives b = 1,169.. = 1,17(to 2 d.p.) The equation of the line of regression is thus: y = −5,07 + 1,17x ˆ 598

A Use the formulae to ﬁnd the regression equation coeﬃcients a and b. B Draw a scatter plot of the data on graph paper. C Now use algebra to ﬁnd a more accurate equation. 2. Footlengths and heights of 7 students are given in the table below. Height (cm) Footlength (cm) 170 27 163 23 131 20 181 28 146 22 134 20 166 24

A Draw a scatter plot of the data on graph paper. B Indentify and describe any trends shown in the scatter plot. C Find the equation of the least squares line by using algebraic methods and draw the line on your graph. D Use your equation to predict the height of a student with footlength 21,6 cm. E Use your equation to predict the footlength of a student 176 cm tall. 3. Repeat the data in question 2 and ﬁnd the regression line using a calculator

44.4.3

Correlation coeﬃcients

Once we have applied regression analysis to a set of data, we would like to have a number that tells us exactly how well the data ﬁts the function. A correlation coeﬃcient, r, is a tool that tells us to what degree there is a relationship between two sets of data. The correlation coeﬃcient r ∈ [−1; 1] when r = −1, there is a perfect negative relationship, when r = 0, there is no relationship and r = 1 is a perfect positive correlation.
y y y y y

x

x

x

x

x

Positive, strong r ≈ 0,9

Positive, fairly strong r ≈ 0,7

Positive, weak r ≈ 0,4

No association r=0

Negative, fairly strong r ≈ −0,7

We often use the correlation coeﬃcient r2 in order to work with the strength of the correlation only (no whether it is positive or negative). In this case: 599

• where n is the number of data points, • sx is the standard deviation of the x-values and • sy is the standard deviation of the y-values. This is known as the Pearson’s product moment correlation coeﬃcient. It is a long calculation and much easier to do on the calculator where you simply follow the procedure for the regression equation, and go on to ﬁnd r.

Draw a scatter plot of the data set and your estimate of a line of best ﬁt. Calculate equation of the line of regression using the method of least squares. Draw the regression line equation onto the graph. Calculate the correlation coeﬃcient r. What conclusion can you reach, regarding the relationship between CO2 emission and GDP per capita for the countries in the data set?

2. A collection of data on the peculiar investigation into a foot size and height of students was recorded in the table below. Answer the questions to follow. Length of right foot (cm) 25,5 26,1 23,7 26,4 27,5 24 22,6 27,1 600 Height (cm) 163,3 164,9 165,5 173,7 174,4 156 155,3 169,3

CHAPTER 44. STATISTICS - GRADE 12 A Draw a scatter plot of the data set and your estimate of a line of best ﬁt.

44.5

B Calculate equation of the line of regression using the method of least squares or your calculator. C Draw the regression line equation onto the graph. D Calculate the correlation coeﬃcient r. E What conclusion can you reach, regarding the relationship between the length of the right foot and height of the students in the data set? 3. A class wrote two tests, and the marks for each were recorded in the table below. Full marks in the ﬁrst test was 50, and the second test was out of 30. A Is there a strong association between the marks for the ﬁrst and second test? Show why or why not. B One of the learners (in row 18) did not write the second test. Given their mark for the ﬁrst test, calculate an expected mark for the second test. Learner 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Test 1 (Full marks: 50) 42 32 31 42 35 23 43 23 24 15 19 13 36 29 29 25 29 17 30 28 Test 2 (Full marks: 30) 25 19 20 26 23 14 24 12 14 10 11 10 22 17 17 16 18 19 17

4. A fast food company produces hamburgers. The number of hamburgers made, and the costs are recorded over a week. Hamburgers made Costs 495 R2382 550 R2442 515 R2484 500 R2400 480 R2370 530 R2448 585 R2805 A Find the linear regression function that best ﬁts the data. B If the total cost in a day is R2500, estimate the number of hamburgers produced. C What is the cost of 490 hamburgers? 5. The proﬁts of a new shop are recorded over the ﬁrst 6 months. The owner wants to predict his future sales. The proﬁts so far have been R90 000 , R93 000, R99 500, R102 000, R101 300, R109 000. A For the proﬁt data, calculate the linear regression function. 601

44.5

CHAPTER 44. STATISTICS - GRADE 12 B Give an estimate of the proﬁts for the next two months. C The owner wants a proﬁt of R130 000. Estimate how many months this will take.

6. A company produces sweets using a machine which runs for a few hours per day. The number of hours running the machine and the number of sweets produced are recorded. Machine hours 3,80 4,23 4,37 4,10 4,17 Sweets produced 275 287 291 281 286

Find the linear regression equation for the data, and estimate the machine hours needed to make 300 sweets.

602

Chapter 45

Combinations and Permutations Grade 12
45.1 Introduction

Mathematics education began with counting. At the beginning, ﬁngers, beans, buttons, and pencils were used to help with counting, but these are only practical for small numbers. What happens when a large number of items must be counted? This chapter focuses on how to use mathematical techniques to count combinations of items.

45.2

Counting

An important aspect of probability theory is the ability to determine the total number of possible outcomes when multiple events are considered. For example, what is the total number of possible outcomes when a die is rolled and then a coin is tossed? The roll of a die has six possible outcomes (1, 2, 3, 4, 5 or 6) and the toss of a coin, 2 outcomes (head or tails). Counting the possible outcomes can be tedious.

The use of lists, tables and tree diagrams is only feasible for events with a few outcomes. When the number of outcomes grows, it is not practical to list the diﬀerent possibilities and the fundamental counting principle is used. The fundamental counting principle describes how to determine the total number of outcomes of a series of events. Suppose that two experiments take place. The ﬁrst experiment has n1 possible outcomes, and the second has n2 possible outcomes. Therefore, the ﬁrst experiment, followed by the second experiment, will have a total of n1 × n2 possible outcomes. This idea can be generalised to m experiments as the total number of outcomes for m experiments is:
m

n1 × n2 × n3 × . . . × nm = is the multiplication equivalent of .

ni
i=1

Note: the order in which the experiments are done does not aﬀect the total number of possible outcomes.

Worked Example 204: Lunch Special Question: A take-away has a 4-piece lunch special which consists of a sandwich, soup, dessert and drink for R25.00. They oﬀer the following choices for : Sandwich: chicken mayonnaise, cheese and tomato, tuna, and ham and lettuce Soup: tomato, chicken noodle, vegetable Dessert: ice-cream, piece of cake Drink: tea, coﬀee, coke, Fanta and Sprite. How many possible meals are there? 604

CHAPTER 45. COMBINATIONS AND PERMUTATIONS - GRADE 12 Answer Step 1 : Determine how many parts to the meal there are There are 4 parts: sandwich, soup, dessert and drink. Step 2 : Identify how many choices there are for each part Meal Component Number of choices Sandwich 4 Soup 3 Dessert 2 Drink 5

The fundamental counting principle describes how to calculate the total number of outcomes when multiple independent events are performed together. A more complex problem is determining how many combinations there are of selecting a group of objects from a set. Mathematically, a combination is deﬁned as an un-ordered collection of unique elements, or more formally, a subset of a set. For example, suppose you have ﬁfty-two playing cards, and select ﬁve cards. The ﬁve cards would form a combination and would be a subset of the set of 52 cards. In a set, the order of the elements in the set does not matter. These are represented usually with curly braces, for example {2, 4, 6} is a subset of the set {1,2,3,4,5,6}. Since the order of the elements does not matter, only the speciﬁc elements are of interest. Therefore, {2, 4, 6} = {6, 4, 2} and {1, 1, 1} is the same as {1} because a set is deﬁned by its elements; they don’t usually appear more than once. Given S, the set of all possible unique elements, a combination is a subset of the elements of S. The order of the elements in a combination is not important (two lists with the same elements in diﬀerent orders are considered to be the same combination). Also, the elements cannot be repeated in a combination (every element appears uniquely once).

45.5.1

Counting Combinations

Calculating the number of ways that certain patterns can be formed is the beginning of combinatorics, the study of combinations. Let S be a set with n objects. Combinations of k objects from this set S are subsets of S having k elements each (where the order of listing the elements does not distinguish two subsets). Combination without Repetition When the order does not matter, but each object can be chosen only once, the number of combinations is: n! n = r!(n − r)! r

where n is the number of objects from which you can choose and r is the number to be chosen. For example, if you have 10 numbers and wish to choose 5 you would have 10!/(5!(10 - 5)!) = 252 ways to choose. For example how many possible 5 card hands are there in a deck of cards with 52 cards? 52! / (5!(52-5)!) = 2 598 960 combinations 605

45.6

CHAPTER 45. COMBINATIONS AND PERMUTATIONS - GRADE 12

Combination with Repetition When the order does not matter and an object can be chosen more than once, then the number of combinations is: (n + r − 1)! n+r−1 n+r−1 = = r!(n − 1)! r n−1

where n is the number of objects from which you can choose and r is the number to be chosen. For example, if you have ten types of donuts to choose from and you want three donuts there are (10 + 3 - 1)! / 3!(10 - 1)! = 220 ways to choose.

45.5.2

Combinatorics and Probability

Combinatorics is quite useful in the computation of probabilities of events, as it can be used to determine exactly how many outcomes are possible in a given event.

Worked Example 205: Probability Question: At a school, learners each play 2 sports. They can choose from netball, basketball, soccer, athletics, swimming, or tennis. What is the probability that a learner plays soccer and either netball, basketball or tennis? Answer Step 1 : Identify what events we are counting We count the events: soccer and netball, soccer and basketball, soccer and tennis. This gives three choices. Step 2 : Calculate the total number of choices There are 6 sports to choose from and we choose 2 sports. There are 6 2 = 6!/(2!(6 − 2)!) = 15 choices. Step 3 : Calculate the probability The probability is the number of events we are counting, divided by the total number of choices. 3 Probability = 15 = 1 = 0,2 5

45.6

Permutations

The concept of a combination did not consider the order of the elements of the subset to be important. A permutation is a combination with the order of a selection from a group being important. For example, for the set {1,2,3,4,5,6}, the combination {1,2,3} would be identical to the combination {3,2,1}, but these two combinations are permutations, because the elements in the set are ordered diﬀerently. More formally, a permutation is an ordered list without repetitions, perhaps missing some elements. This means that {1, 2, 2, 3, 4, 5, 6} and {1, 2, 4, 5, 5, 6} are not permutations of the set {1, 2, 3, 4, 5, 6}. Now suppose you have these objects: 1, 2, 3 Here is a list of all permutations of those: 1 2 3; 1 3 2; 2 1 3; 2 3 1; 3 1 2; 3 2 1; 606

CHAPTER 45. COMBINATIONS AND PERMUTATIONS - GRADE 12

45.6

45.6.1

Counting Permutations

Let S be a set with n objects. Permutations of k objects from this set S refer to sequences of k diﬀerent elements of S (where two sequences are considered diﬀerent if they contain the same elements but in a diﬀerent order, or if they have a diﬀerent length). Formulas for the number of permutations and combinations are readily available and important throughout combinatorics. It is easy to count the number of permutations of size r when chosen from a set of size n (with r ≤ n). 1. Select the ﬁrst member of all permutations out of n choices because there are n distinct elements in the set. 2. Next, since one of the n elements has already been used, the second member of the permutation has (n − 1) elements to choose from the remaining set. 3. The third member of the permutation can be ﬁlled in (n − 2) ways since 2 have been used already. 4. This pattern continues until there are r members on the permutation. This means that the last member can be ﬁlled in (n − (r − 1)) = (n − r + 1) ways. 5. Summarizing, we ﬁnd that there is a total of n(n − 1)(n − 2)...(n − r + 1) diﬀerent permutations of r objects, taken from a pool of n objects. This number is denoted by P (n, r) and can be written in factorial notation as: P (n,r) = n! . (n − r)!

For example, if we have a total of 5 elements, the integers {1, 2, 3,4,5}, how many ways are there for a permutation of three elements to be selected from this set? In this case, n = 10 and r = 3. Then, P (10,3) = 10!/7! = 720.

Worked Example 206: Permutations Question: Show that a collection of n objects has n! permutations. Answer Proof: Constructing an ordered sequence of n objects is equivalent to choosing the position occupied by the ﬁrst object, then choosing the position of the second object, and so on, until we have chosen the position of each of our n objects. There are n ways to choose a position for the ﬁrst object. Once its position is ﬁxed, we can choose from (n-1) possible positions for the second object. With the ﬁrst two placed, there are (n-2) remaining possible positions for the third object; and so on. There are only two positions to choose from for the penultimate object, and the nth object will occupy the last remaining position. Therefore, according to the multiplicative principle, there are n(n − 1)(n − 2)...2 × 1 = n! ways of constructing an ordered sequence of n objects.

607

45.7

CHAPTER 45. COMBINATIONS AND PERMUTATIONS - GRADE 12

Permutation with Repetition When order matters and an object can be chosen more than once then the number of permutations is: nr where n is the number of objects from which you can choose and r is the number to be chosen. For example, if you have the letters A, B, C, and D and you wish to discover the number of ways of arranging them in three letter patterns (trigrams) you ﬁnd that there are 43 or 64 ways. This is because for the ﬁrst slot you can choose any of the four values, for the second slot you can choose any of the four, and for the ﬁnal slot you can choose any of the four letters. Multiplying them together gives the total. Permutation without Repetition When the order matters and each object can be chosen only once, then the number of permutations is: n! (n − r)!

where n is the number of objects from which you can choose and r is the number to be chosen.

For example, if you have ﬁve people and are going to choose three out of these, you will have 5!/(5-3)! = 60 permutations. Note that if n = r (meaning number of chosen elements is equal to number of elements to choose from) then the formula becomes n! n! = = n! (n − n)! 0! For example, if you have three people and you want to ﬁnd out how many ways you may arrange them it would be 3! or 3 × 2 × 1 = 6 ways. The reason for this is because you can choose from three for the initial slot, then you are left with only two to choose from for the second slot, and that leaves only one for the ﬁnal slot. Multiplying them together gives the total.

45.7

Applications

Extension: The Binomial Theorem In mathematics, the binomial theorem is an important formula giving the expansion of powers of sums. Its simplest version reads
n

The coeﬃcients form a triangle, where each number is the sum of the two numbers above it:
1

1 1 1 4 3 6 2

1 1 3 4 1 1

This formula, and the triangular arrangement of the binomial coeﬃcients, are often attributed to Blaise Pascal who described them in the 17th century. It was, however, known to the Chinese mathematician Yang Hui in the 13th century, the earlier Persian mathematician Omar Khayym in the 11th century, and the even earlier Indian mathematician Pingala in the 3rd century BC.

Worked Example 207: Number Plates Question: The number plate on a car consists of any 3 letters of the alphabet (excluding the vowels and ’Q’), followed by any 3 digits (0 to 9). For a car chosen at random, what is the probability that the number plate starts with a ’Y’ and ends with an odd digit? Answer Step 1 : Identify what events are counted The number plate starts with a ’Y’, so there is only 1 choice for the ﬁrst letter, and ends with an even digit, so there are 5 choices for the last digit (1,3,5,7,9). Step 2 : Find the number of events Use the counting principle. For each of the other letters, there are 20 possible choices (26 in the alphabet, minus 5 vowels and ’Q’) and 10 possible choices for each of the other digits. Number of events = 1 × 20 × 20 × 10 × 10 × 5 = 200 000 Step 3 : Find the number of total possible number plates Use the counting principle. This time, the ﬁrst letter and last digit can be anything. Total number of choices = 20 × 20 × 20 × 10 × 10 × 10 = 8 000 000 Step 4 : Calculate the probability The probability is the number of events we are counting, divided by the total number of choices. 1 Probability = 8200 000 = 40 = 0,025 000 000

1. Tshepo and Sally go to a restaurant, where the menu is: Starter Main Course Dessert Chicken wings Beef burger Chocolate ice cream Mushroom soup Chicken burger Strawberry ice cream Greek salad Chicken curry Apple crumble Lamb curry Chocolate mousse Vegetable lasagne A How many diﬀerent combinations (of starter, main meal, and dessert) can Tshepo have? B Sally doesn’t like chicken. How many diﬀerent combinations can she have? 2. Four coins are thrown, and the outcomes recorded. How many diﬀerent ways are there of getting three heads? First write out the possibilites, and then use the formula for combinations. 3. The answers in a multiple choice test can be A, B, C, D, or E. In a test of 12 questions, how many diﬀerent ways are there of answering the test? 4. A girl has 4 dresses, 2 necklaces, and 3 handbags. A How many diﬀerent choices of outﬁt (dress, necklace and handbag) does she have? B She now buys two pairs of shoes. How many choices of outﬁt (dress, necklace, handbag and shoes) does she now have? 5. In a soccer tournament of 9 teams, every team plays every other team. A How many matches are there in the tournament? B If there are 5 boys’ teams and 4 girls’ teams, what is the probability that the ﬁrst match will be played between 2 girls’ teams? 6. The letters of the word ’BLUE’ are rearranged randomly. How many new words (a word is any combination of letters) can be made? 7. The letters of the word ’CHEMISTRY’ are arranged randomly to form a new word. What is the probability that the word will start and end with a vowel? 610

CHAPTER 45. COMBINATIONS AND PERMUTATIONS - GRADE 12

45.8

8. There are 2 History classes, 5 Accounting classes, and 4 Mathematics classes at school. Luke wants to do all three subjects. How many possible combinations of classes are there? 9. A school netball team has 8 members. How many ways are there to choose a captain, vice-captain, and reserve? 10. A class has 15 boys and 10 girls. A debating team of 4 boys and 6 girls must be chosen. How many ways can this be done? 11. A secret pin number is 3 characters long, and can use any digit (0 to 9) or any letter of the alphabet. Repeated characters are allowed. How many possible combinations are there?

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COPYING IN QUANTITY
If you publish printed copies (or copies in media that commonly have printed covers) of the Document, numbering more than 100, and the Document’s license notice requires Cover Texts, 620

APPENDIX A. GNU FREE DOCUMENTATION LICENSE you must enclose the copies in covers that carry, clearly and legibly, all these Cover Texts: FrontCover Texts on the front cover, and Back-Cover Texts on the back cover. Both covers must also clearly and legibly identify you as the publisher of these copies. The front cover must present the full title with all words of the title equally prominent and visible. You may add other material on the covers in addition. Copying with changes limited to the covers, as long as they preserve the title of the Document and satisfy these conditions, can be treated as verbatim copying in other respects. If the required texts for either cover are too voluminous to ﬁt legibly, you should put the ﬁrst ones listed (as many as ﬁt reasonably) on the actual cover, and continue the rest onto adjacent pages. If you publish or distribute Opaque copies of the Document numbering more than 100, you must either include a machine-readable Transparent copy along with each Opaque copy, or state in or with each Opaque copy a computer-network location from which the general network-using public has access to download using public-standard network protocols a complete Transparent copy of the Document, free of added material. If you use the latter option, you must take reasonably prudent steps, when you begin distribution of Opaque copies in quantity, to ensure that this Transparent copy will remain thus accessible at the stated location until at least one year after the last time you distribute an Opaque copy (directly or through your agents or retailers) of that edition to the public. It is requested, but not required, that you contact the authors of the Document well before redistributing any large number of copies, to give them a chance to provide you with an updated version of the Document.

MODIFICATIONS
You may copy and distribute a Modiﬁed Version of the Document under the conditions of sections A and A above, provided that you release the Modiﬁed Version under precisely this License, with the Modiﬁed Version ﬁlling the role of the Document, thus licensing distribution and modiﬁcation of the Modiﬁed Version to whoever possesses a copy of it. In addition, you must do these things in the Modiﬁed Version: 1. Use in the Title Page (and on the covers, if any) a title distinct from that of the Document, and from those of previous versions (which should, if there were any, be listed in the History section of the Document). You may use the same title as a previous version if the original publisher of that version gives permission. 2. List on the Title Page, as authors, one or more persons or entities responsible for authorship of the modiﬁcations in the Modiﬁed Version, together with at least ﬁve of the principal authors of the Document (all of its principal authors, if it has fewer than ﬁve), unless they release you from this requirement. 3. State on the Title page the name of the publisher of the Modiﬁed Version, as the publisher. 4. Preserve all the copyright notices of the Document. 5. Add an appropriate copyright notice for your modiﬁcations adjacent to the other copyright notices. 6. Include, immediately after the copyright notices, a license notice giving the public permission to use the Modiﬁed Version under the terms of this License, in the form shown in the Addendum below. 7. Preserve in that license notice the full lists of Invariant Sections and required Cover Texts given in the Document’s license notice. 8. Include an unaltered copy of this License. 9. Preserve the section Entitled “History”, Preserve its Title, and add to it an item stating at least the title, year, new authors, and publisher of the Modiﬁed Version as given on the Title Page. If there is no section Entitled “History” in the Document, create one stating the title, year, authors, and publisher of the Document as given on its Title Page, then add an item describing the Modiﬁed Version as stated in the previous sentence. 621

APPENDIX A. GNU FREE DOCUMENTATION LICENSE 10. Preserve the network location, if any, given in the Document for public access to a Transparent copy of the Document, and likewise the network locations given in the Document for previous versions it was based on. These may be placed in the “History” section. You may omit a network location for a work that was published at least four years before the Document itself, or if the original publisher of the version it refers to gives permission. 11. For any section Entitled “Acknowledgements” or “Dedications”, Preserve the Title of the section, and preserve in the section all the substance and tone of each of the contributor acknowledgements and/or dedications given therein. 12. Preserve all the Invariant Sections of the Document, unaltered in their text and in their titles. Section numbers or the equivalent are not considered part of the section titles. 13. Delete any section Entitled “Endorsements”. Such a section may not be included in the Modiﬁed Version. 14. Do not re-title any existing section to be Entitled “Endorsements” or to conﬂict in title with any Invariant Section. 15. Preserve any Warranty Disclaimers. If the Modiﬁed Version includes new front-matter sections or appendices that qualify as Secondary Sections and contain no material copied from the Document, you may at your option designate some or all of these sections as invariant. To do this, add their titles to the list of Invariant Sections in the Modiﬁed Version’s license notice. These titles must be distinct from any other section titles. You may add a section Entitled “Endorsements”, provided it contains nothing but endorsements of your Modiﬁed Version by various parties–for example, statements of peer review or that the text has been approved by an organisation as the authoritative deﬁnition of a standard. You may add a passage of up to ﬁve words as a Front-Cover Text, and a passage of up to 25 words as a Back-Cover Text, to the end of the list of Cover Texts in the Modiﬁed Version. Only one passage of Front-Cover Text and one of Back-Cover Text may be added by (or through arrangements made by) any one entity. If the Document already includes a cover text for the same cover, previously added by you or by arrangement made by the same entity you are acting on behalf of, you may not add another; but you may replace the old one, on explicit permission from the previous publisher that added the old one. The author(s) and publisher(s) of the Document do not by this License give permission to use their names for publicity for or to assert or imply endorsement of any Modiﬁed Version.

COMBINING DOCUMENTS
You may combine the Document with other documents released under this License, under the terms deﬁned in section A above for modiﬁed versions, provided that you include in the combination all of the Invariant Sections of all of the original documents, unmodiﬁed, and list them all as Invariant Sections of your combined work in its license notice, and that you preserve all their Warranty Disclaimers. The combined work need only contain one copy of this License, and multiple identical Invariant Sections may be replaced with a single copy. If there are multiple Invariant Sections with the same name but diﬀerent contents, make the title of each such section unique by adding at the end of it, in parentheses, the name of the original author or publisher of that section if known, or else a unique number. Make the same adjustment to the section titles in the list of Invariant Sections in the license notice of the combined work. In the combination, you must combine any sections Entitled “History” in the various original documents, forming one section Entitled “History”; likewise combine any sections Entitled “Acknowledgements”, and any sections Entitled “Dedications”. You must delete all sections Entitled “Endorsements”. 622

APPENDIX A. GNU FREE DOCUMENTATION LICENSE

COLLECTIONS OF DOCUMENTS
You may make a collection consisting of the Document and other documents released under this License, and replace the individual copies of this License in the various documents with a single copy that is included in the collection, provided that you follow the rules of this License for verbatim copying of each of the documents in all other respects. You may extract a single document from such a collection, and distribute it individually under this License, provided you insert a copy of this License into the extracted document, and follow this License in all other respects regarding verbatim copying of that document.

AGGREGATION WITH INDEPENDENT WORKS
A compilation of the Document or its derivatives with other separate and independent documents or works, in or on a volume of a storage or distribution medium, is called an “aggregate” if the copyright resulting from the compilation is not used to limit the legal rights of the compilation’s users beyond what the individual works permit. When the Document is included an aggregate, this License does not apply to the other works in the aggregate which are not themselves derivative works of the Document. If the Cover Text requirement of section A is applicable to these copies of the Document, then if the Document is less than one half of the entire aggregate, the Document’s Cover Texts may be placed on covers that bracket the Document within the aggregate, or the electronic equivalent of covers if the Document is in electronic form. Otherwise they must appear on printed covers that bracket the whole aggregate.

TRANSLATION
Translation is considered a kind of modiﬁcation, so you may distribute translations of the Document under the terms of section A. Replacing Invariant Sections with translations requires special permission from their copyright holders, but you may include translations of some or all Invariant Sections in addition to the original versions of these Invariant Sections. You may include a translation of this License, and all the license notices in the Document, and any Warranty Disclaimers, provided that you also include the original English version of this License and the original versions of those notices and disclaimers. In case of a disagreement between the translation and the original version of this License or a notice or disclaimer, the original version will prevail. If a section in the Document is Entitled “Acknowledgements”, “Dedications”, or “History”, the requirement (section A) to Preserve its Title (section A) will typically require changing the actual title.

TERMINATION
You may not copy, modify, sub-license, or distribute the Document except as expressly provided for under this License. Any other attempt to copy, modify, sub-license or distribute the Document is void, and will automatically terminate your rights under this License. However, parties who have received copies, or rights, from you under this License will not have their licenses terminated so long as such parties remain in full compliance.

FUTURE REVISIONS OF THIS LICENSE
The Free Software Foundation may publish new, revised versions of the GNU Free Documentation License from time to time. Such new versions will be similar in spirit to the present version, but may diﬀer in detail to address new problems or concerns. See http://www.gnu.org/copyleft/. 623

APPENDIX A. GNU FREE DOCUMENTATION LICENSE Each version of the License is given a distinguishing version number. If the Document speciﬁes that a particular numbered version of this License “or any later version” applies to it, you have the option of following the terms and conditions either of that speciﬁed version or of any later version that has been published (not as a draft) by the Free Software Foundation. If the Document does not specify a version number of this License, you may choose any version ever published (not as a draft) by the Free Software Foundation.

ADDENDUM: How to use this License for your documents
To use this License in a document you have written, include a copy of the License in the document and put the following copyright and license notices just after the title page: Copyright c YEAR YOUR NAME. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled “GNU Free Documentation License”. If you have Invariant Sections, Front-Cover Texts and Back-Cover Texts, replace the “with...Texts.” line with this: with the Invariant Sections being LIST THEIR TITLES, with the Front-Cover Texts being LIST, and with the Back-Cover Texts being LIST. If you have Invariant Sections without Cover Texts, or some other combination of the three, merge those two alternatives to suit the situation. If your document contains nontrivial examples of program code, we recommend releasing these examples in parallel under your choice of free software license, such as the GNU General Public License, to permit their use in free software.